Exercise 1: Four limits, and the algebra that opens each one

5 answers to check

Evaluate each limit exactly, or say that it is infinite. L'Hôpital's Rule is NOT allowed in this question: the first line of each part names the algebraic move (factor, conjugate, match the angle, squeeze).

a) $\displaystyle\lim_{x\to 7} \frac{x^2 - 5x - 14}{\sqrt{x + 2} - 3}$
b) $\displaystyle\lim_{x\to 0} \frac{\sin(x^2)}{1 - \cos x}$
c) $\displaystyle\lim_{x\to -\infty} \left(\sqrt{9x^2 + 6x} + 3x\right)$ and $\displaystyle\lim_{x\to \infty} \left(\sqrt{9x^2 + 6x} + 3x\right)$
d) $\displaystyle\lim_{x\to \infty} \frac{5x + 2\sin x}{\sqrt{x^2 + 1}}$

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Answers

a) $54$
b) $2$
c) $-1$ at $-\infty$ ; $+\infty$ at $+\infty$ (no indeterminate form there)
d) $5$ : $\frac{\sin x}{x} \to 0$ at $\infty$ , by the Sandwich Theorem

a)

Substitution gives $\frac{49 - 35 - 14}{3 - 3} = \frac{0}{0}$ : the top and the bottom both hide the factor $x - 7$ . Factor the trinomial, $x^2 - 5x - 14 = (x - 7)(x + 2)$ , and multiply top AND bottom by the conjugate $\sqrt{x + 2} + 3$ : the denominator becomes $(x + 2) - 9 = x - 7$ . For $x \ne 7$ the quotient is $\frac{(x - 7)(x + 2)\left(\sqrt{x + 2} + 3\right)}{x - 7} = (x + 2)\left(\sqrt{x + 2} + 3\right)$ , continuous at $7$ , so the limit is $9 \cdot 6 = 54$ . Multiplying the denominator alone by the conjugate changes the value of the fraction: both floors, always.

b)

Substitution gives $\frac{\sin 0}{1 - 1} = \frac{0}{0}$ . Two known limits are hidden here, and the algebra must bring each one out. On top, the angle is $x^2$ , so divide by $x^2$ : $\frac{\sin(x^2)}{x^2} \to 1$ as $x^2 \to 0$ . Below, the half angle identity $1 - \cos x = 2\sin^2\frac{x}{2}$ gives $\frac{1 - \cos x}{x^2} = \frac{2\sin^2(x/2)}{x^2} = \frac{1}{2}\left(\frac{\sin(x/2)}{x/2}\right)^2 \to \frac{1}{2}$ . So $\frac{\sin(x^2)}{1 - \cos x} = \frac{\sin(x^2)}{x^2} \cdot \frac{x^2}{1 - \cos x} \to 1 \cdot 2 = 2$ . The factor $\frac{x^2}{x^2}$ is exactly what was multiplied in to match each angle with its denominator; it must be written, not guessed.

c)

At $+\infty$ both terms tend to $+\infty$ and so does their SUM: no indeterminate form, no algebra, the limit is $+\infty$ . At $-\infty$ , $\sqrt{9x^2 + 6x} \to +\infty$ while $3x \to -\infty$ : the form $\infty - \infty$ . Multiply and divide by the conjugate $\sqrt{9x^2 + 6x} - 3x$ : $\frac{(9x^2 + 6x) - 9x^2}{\sqrt{9x^2 + 6x} - 3x} = \frac{6x}{\sqrt{9x^2 + 6x} - 3x}$ . Divide top and bottom by $x$ . For $x < 0$ , $\sqrt{x^2} = |x| = -x$ , so $\frac{\sqrt{9x^2 + 6x}}{x} = -\sqrt{9 + \frac{6}{x}}$ , and the quotient is $\frac{6}{-\sqrt{9 + 6/x} - 3} \to \frac{6}{-6} = -1$ . Writing $\sqrt{x^2} = x$ here gives $\frac{6}{3 - 3}$ , a division by zero that should have rung the alarm.

d)

Move: divide top and bottom by $x$ , the dominant power, with $\sqrt{x^2} = |x| = x$ for $x > 0$ : the quotient becomes $\frac{5 + \frac{2\sin x}{x}}{\sqrt{1 + \frac{1}{x^2}}}$ . The quotient law needs the limit of $\frac{\sin x}{x}$ at infinity, and $\sin x$ itself has none: it oscillates forever. But it is bounded, and that is enough: for $x > 0$ , $-\frac{1}{x} \le \frac{\sin x}{x} \le \frac{1}{x}$ , both bounds tend to $0$ , so by the Sandwich Theorem $\frac{\sin x}{x} \to 0$ . Not $1$ : the limit $\frac{\sin\theta}{\theta} \to 1$ is at $0$ , not at infinity. Now every piece has a limit and the quotient law applies: $\frac{5 + 0}{\sqrt{1 + 0}} = 5$ .

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Exercise 2: Three derivatives: every rule named, every result simplified

8 answers to check

Name each rule as you use it (product, quotient, chain with its inner function written down), then simplify: the factored form is part of the answer.

a) Let $f(x) = \dfrac{(x + 1)^3}{\sqrt{2x + 3}}$ . Give its domain, write $f'(x)$ as one factored fraction, compute $f'(3)$ , and find every point where the tangent is horizontal.
b) Let $g(x) = \arcsin\left(\dfrac{x - 1}{x + 1}\right)$ . Find the domain of $g$ , show that $g'(x) = \dfrac{1}{(x + 1)\sqrt{x}}$ for $x > 0$ , and compute $g'(1)$ and $g'(4)$ .
c) Let $y = xe^{-2x}$ . Find $y'$ and $y''$ , factoring out $e^{-2x}$ at each step, and show that $y'' + 4y' + 4y = 0$ for every $x$ . Give $y'(0)$ and $y''(0)$ .

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Answers

a) Domain $\left(-\frac{3}{2}, \infty\right)$ ; $f'(x) = \frac{(x + 1)^2(5x + 8)}{(2x + 3)^{3/2}}$ ; $f'(3) = \frac{368}{27}$ ; one horizontal tangent, at $(-1, 0)$ ( $x = -\frac{8}{5}$ rejected)
b) Domain $[0, \infty)$ ; $g'(1) = \frac{1}{2}$ , $g'(4) = \frac{1}{10}$
c) $y' = e^{-2x}(1 - 2x)$ , $y'' = 4e^{-2x}(x - 1)$ ; $y'(0) = 1$ , $y''(0) = -4$

a)

The root needs $2x + 3 > 0$ (it is also a denominator): the domain is $\left(-\frac{3}{2}, \infty\right)$ . Rewrite the root as a power before differentiating: $f(x) = (x + 1)^3(2x + 3)^{-1/2}$ . Product rule, with the chain rule on each factor (inner functions $x + 1$ and $2x + 3$ ): $f'(x) = 3(x + 1)^2(2x + 3)^{-1/2} + (x + 1)^3 \cdot \left(-\frac{1}{2}\right)(2x + 3)^{-3/2} \cdot 2$ . Factor the LOWEST power of each bracket, $(x + 1)^2$ and $(2x + 3)^{-3/2}$ : $f'(x) = (x + 1)^2(2x + 3)^{-3/2}\left[3(2x + 3) - (x + 1)\right] = \frac{(x + 1)^2(5x + 8)}{(2x + 3)^{3/2}}$ . The $2x + 3$ inside the bracket is what is left of $(2x + 3)^{-1/2}$ after taking out $(2x + 3)^{-3/2}$ : the exponents subtract, $-\frac{1}{2} - \left(-\frac{3}{2}\right) = 1$ .

At $x = 3$ : $f'(3) = \frac{16 \cdot 23}{9^{3/2}} = \frac{368}{27} \approx 13.6296$ , since $9^{3/2} = 27$ . Horizontal tangents: the numerator vanishes at $x = -1$ and at $x = -\frac{8}{5}$ . But $-\frac{8}{5} = -1.6 < -1.5$ lies OUTSIDE the domain: rejected. There is exactly one horizontal tangent, at $(-1, 0)$ .

b)

Domain first. $\arcsin u$ needs $-1 \le u \le 1$ , with $u = \frac{x - 1}{x + 1}$ . For $x > -1$ the denominator is positive: $u \le 1$ reads $x - 1 \le x + 1$ , always true, and $u \ge -1$ reads $x - 1 \ge -(x + 1)$ , that is $x \ge 0$ . For $x < -1$ , $u = 1 + \frac{-2}{x + 1} > 1$ . The domain is $[0, \infty)$ . Chain rule with the inner function $u$ : $g'(x) = \frac{u'}{\sqrt{1 - u^2}}$ , and the quotient rule gives $u' = \frac{(x + 1) - (x - 1)}{(x + 1)^2} = \frac{2}{(x + 1)^2}$ .

The algebra is under the root. Bring $1 - u^2$ to one fraction and use a DIFFERENCE OF SQUARES instead of expanding: $1 - u^2 = \frac{(x + 1)^2 - (x - 1)^2}{(x + 1)^2} = \frac{\left[(x + 1) - (x - 1)\right]\left[(x + 1) + (x - 1)\right]}{(x + 1)^2} = \frac{4x}{(x + 1)^2}$ . Since $x + 1 > 0$ , $\sqrt{1 - u^2} = \frac{2\sqrt{x}}{x + 1}$ . Then $g'(x) = \frac{2}{(x + 1)^2} \cdot \frac{x + 1}{2\sqrt{x}} = \frac{1}{(x + 1)\sqrt{x}}$ for $x > 0$ . So $g'(1) = \frac{1}{2}$ and $g'(4) = \frac{1}{5 \cdot 2} = \frac{1}{10}$ . At $x = 0$ , $u = -1$ and $g'(x) \to \infty$ : the graph starts with a vertical tangent.

c)

Product rule, with the chain rule on $e^{-2x}$ (inner function $-2x$ , derivative $-2$ ): $y' = e^{-2x} + x(-2)e^{-2x} = e^{-2x}(1 - 2x)$ . Again, product rule: $y'' = -2e^{-2x}(1 - 2x) + e^{-2x}(-2) = e^{-2x}(-2 + 4x - 2) = 4e^{-2x}(x - 1)$ . Keeping $e^{-2x}$ factored, never distributed, is what makes the check one line: $y'' + 4y' + 4y = e^{-2x}\left[(4x - 4) + 4(1 - 2x) + 4x\right] = e^{-2x} \cdot 0 = 0$ for every $x$ . Values: $y'(0) = 1$ and $y''(0) = -4$ .

Exercise 3: A moving exponent, and the slope of an inverse with no formula

3 answers to check

Parts a) and b) are independent. Three points each.

a) Let $y = \left(1 + \dfrac{1}{x}\right)^x$ for $x > 0$ . Explain why neither the power rule nor the rule for $a^x$ applies, find $\frac{y'}{y}$ by logarithmic differentiation, expanding with the laws of logarithms first, and compute $y'(1)$ .
b) Let $f(x) = 3x + \sin x$ . Explain why $f$ is one-to-one. Find $(f^{-1})'(0)$ and the tangent line to $y = f^{-1}(x)$ at the origin, then $(f^{-1})'\left(\frac{3\pi}{2} + 1\right)$ .

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Answers

a) $\frac{y'}{y} = \ln\frac{x + 1}{x} - \frac{1}{x + 1}$ ; $y'(1) = 2\ln 2 - 1 \approx 0.3863$
b) $f' \ge 2 > 0$ ; $(f^{-1})'(0) = \frac{1}{4}$ , tangent $y = \frac{x}{4}$ ; $(f^{-1})'\left(\frac{3\pi}{2} + 1\right) = \frac{1}{3}$

a)

The variable sits in the base AND in the exponent: the power rule needs a constant exponent, the rule for $a^x$ a constant base. Take logarithms, $y > 0$ : $\ln y = x\ln\left(1 + \frac{1}{x}\right)$ . Before differentiating, write the inside as ONE fraction and split it with the quotient law: $1 + \frac{1}{x} = \frac{x + 1}{x}$ , so $\ln y = x\ln(x + 1) - x\ln x$ . Differentiate, the product rule on each term: $\frac{y'}{y} = \ln(x + 1) + \frac{x}{x + 1} - \ln x - 1 = \ln\frac{x + 1}{x} - \frac{1}{x + 1}$ , since $\frac{x}{x + 1} - 1 = -\frac{1}{x + 1}$ . At $x = 1$ : $y(1) = 2$ and $\frac{y'}{y} = \ln 2 - \frac{1}{2}$ , so $y'(1) = 2\ln 2 - 1 \approx 0.3863$ . Forgetting to multiply by $y(1) = 2$ gives half the answer.

b)

$f'(x) = 3 + \cos x \ge 3 - 1 = 2 > 0$ for every $x$ : $f$ is increasing on $\mathbb{R}$ , so it never takes the same value twice, and it is one-to-one. No formula for $f^{-1}$ exists, and none is needed: if $f(a) = b$ , then $(f^{-1})'(b) = \frac{1}{f'(a)}$ . For $b = 0$ : $f(0) = 0$ , so $a = 0$ and $(f^{-1})'(0) = \frac{1}{3 + 1} = \frac{1}{4}$ . The tangent to $y = f^{-1}(x)$ at the origin is $y = \frac{x}{4}$ , the reflection in $y = x$ of the tangent $y = 4x$ of $f$ .

For $b = \frac{3\pi}{2} + 1$ , find $a$ FIRST: $f\left(\frac{\pi}{2}\right) = \frac{3\pi}{2} + 1$ , so $a = \frac{\pi}{2}$ and $(f^{-1})'(b) = \frac{1}{3 + \cos\frac{\pi}{2}} = \frac{1}{3}$ . Evaluating $f'$ at $b$ itself gives $\frac{1}{3 + \cos\left(\frac{3\pi}{2} + 1\right)} = \frac{1}{3 + \sin 1} \approx 0.26$ : the slope of the wrong point.

Exercise 4: Two seams to close, then a root located by bisection

5 answers to check

Parts a) and b) are independent. In a), $a$ and $b$ are constants and $h(x) = \dfrac{\sin 3x}{\tan x}$ for $-\frac{\pi}{2} < x < 0$ , $h(x) = ax + b$ for $0 \le x \le 1$ , and $h(x) = \dfrac{x - x^{-1}}{x - 1}$ for $x > 1$ . The figure belongs to b), where a calculator is allowed; give its values to four decimals.

a) Find $a$ and $b$ such that $h$ is continuous on $\left(-\frac{\pi}{2}, \infty\right)$ . Each one-sided limit needs one algebraic move: name it.
b) Prove that the equation $\ln x = 3 - x$ has exactly one solution, and that it lies in $(2, 3)$ , naming the function, the interval and each hypothesis of the Intermediate Value Theorem. Then do four steps of bisection on $[2, 3]$ with your calculator, in a table, and give the final interval, of length $\frac{1}{16}$ .

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Answers

a) Left limit at $0$ : $3$ , so $b = 3$ ; right limit at $1$ : $2$ , so $a + b = 2$ and $a = -1$
b) $g(x) = \ln x + x - 3$ : $g(2) < 0 < g(3)$ , $g' > 0$ ; the root lies in $[2.1875, 2.25]$ (about $2.2079$ )

a)

On each open piece $h$ is continuous: $\frac{\sin 3x}{\tan x}$ is a quotient of continuous functions whose denominator does not vanish on $\left(-\frac{\pi}{2}, 0\right)$ , the middle piece is a polynomial, and $\frac{x - x^{-1}}{x - 1}$ has a nonzero denominator for $x > 1$ . Only the seams remain, one equation each. At $0$ , the middle formula owns the point: $h(0) = b$ , and the right-hand limit is $b$ . Left-hand limit, form $\frac{0}{0}$ ; move: write $\tan x = \frac{\sin x}{\cos x}$ and match each sine with its angle, $\frac{\sin 3x}{\tan x} = \frac{\sin 3x\cos x}{\sin x} = 3 \cdot \frac{\sin 3x}{3x} \cdot \frac{x}{\sin x} \cdot \cos x \to 3 \cdot 1 \cdot 1 \cdot 1 = 3$ . Continuity at $0$ requires $b = 3$ .

At $1$ : $h(1) = a + b$ , which is also the left-hand limit. Right-hand limit, form $\frac{0}{0}$ ; move: rewrite the NEGATIVE exponent first, $x - x^{-1} = x - \frac{1}{x} = \frac{x^2 - 1}{x}$ , then factor the difference of squares: for $x > 1$ , $h(x) = \frac{(x - 1)(x + 1)}{x(x - 1)} = \frac{x + 1}{x} \to 2$ . Continuity at $1$ requires $a + b = 2$ , so $a = -1$ . With $a = -1$ and $b = 3$ , the middle piece $3 - x$ joins $3$ on the left and $2$ on the right.

b)

Let $g(x) = \ln x + x - 3$ , so that the equation reads $g(x) = 0$ . Existence: $g$ is continuous on $[2, 3]$ (a logarithm, continuous for $x > 0$ , plus a polynomial); $g(2) = \ln 2 - 1 \approx -0.3069 < 0$ and $g(3) = \ln 3 \approx 1.0986 > 0$ . By the Intermediate Value Theorem, $g(c) = 0$ for some $c$ in $(2, 3)$ . Uniqueness: $g'(x) = \frac{1}{x} + 1 > 0$ for $x > 0$ , so $g$ is increasing on $(0, \infty)$ and takes the value $0$ only once. The figure shows it: $\ln x$ rises, $3 - x$ falls, they cross once.

Bisection, keeping at each step the half where $g$ changes sign: $g(2.5) \approx 0.4163 > 0$ , keep $[2, 2.5]$ ; $g(2.25) \approx 0.0609 > 0$ , keep $[2, 2.25]$ ; $g(2.125) \approx -0.1212 < 0$ , keep $[2.125, 2.25]$ ; $g(2.1875) \approx -0.0297 < 0$ , keep $[2.1875, 2.25]$ . The final interval $[2.1875, 2.25]$ has length $\frac{1}{16}$ and contains the solution, which is about $2.2079$ . Each step needs the SIGN only, but a sign read from a value rounded too hard, $g(2.25) \approx 0.1$ or $0.0$ , is where the table goes wrong.

Exercise 5: The derivative from its definition, then read backwards

5 answers to check

Let $f(x) = \dfrac{3}{x^2 + 2}$ . In a), the derivative must come from the definition $f'(x) = \lim_{h\to 0}\frac{f(x + h) - f(x)}{h}$ ; the differentiation rules may only check the answer.

a) Find $f'(x)$ from the definition, then the tangent line to the graph of $f$ at $x = 1$ .
b) Each limit is a derivative $F'(c)$ : name $F$ and $c$ , then evaluate.(i) $\displaystyle\lim_{h\to 0}\frac{e^{\sin h} - 1}{h}$ (ii) $\displaystyle\lim_{x\to \pi/6}\frac{2\sin x - 1}{6x - \pi}$ .
c) With the function $f$ of a), find $\displaystyle\lim_{h\to 0}\frac{f(1 + 2h) - f(1 - h)}{h}$ without computing any new difference quotient.

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Answers

a) $f'(x) = -\frac{6x}{(x^2 + 2)^2}$ ; tangent $y = 1 - \frac{2}{3}(x - 1)$
b) (i) $F(x) = e^{\sin x}$ , $c = 0$ : $1$ ; (ii) $F = \sin$ , $c = \frac{\pi}{6}$ , times $\frac{1}{3}$ : $\frac{\sqrt{3}}{6}$
c) $3f'(1) = -2$

a)

Numerator of the difference quotient over ONE common denominator: $f(x + h) - f(x) = \frac{3(x^2 + 2) - 3\left((x + h)^2 + 2\right)}{\left((x + h)^2 + 2\right)(x^2 + 2)} = \frac{3\left(x^2 - (x + h)^2\right)}{\left((x + h)^2 + 2\right)(x^2 + 2)}$ . The difference of squares $x^2 - (x + h)^2 = -h(2x + h)$ shows the factor $h$ , as it must: a numerator that is not a multiple of $h$ signals an algebra error. Dividing by $h$ cancels it: $\frac{f(x + h) - f(x)}{h} = \frac{-3(2x + h)}{\left((x + h)^2 + 2\right)(x^2 + 2)} \to \frac{-6x}{(x^2 + 2)^2}$ . At $x = 1$ : $f(1) = 1$ and $f'(1) = -\frac{6}{9} = -\frac{2}{3}$ , so the tangent is $y = 1 - \frac{2}{3}(x - 1)$ . Check with the rules: $f = 3(x^2 + 2)^{-1}$ gives $f' = -3(x^2 + 2)^{-2} \cdot 2x$ .

b)

(i) The shape is $\frac{F(c + h) - F(c)}{h}$ with $F(x) = e^{\sin x}$ and $c = 0$ : the subtracted number $1$ is exactly $F(0) = e^{\sin 0} = e^0$ . The limit is $F'(0)$ , and the chain rule with the inner function $\sin x$ gives $F'(x) = e^{\sin x}\cos x$ , so the limit is $F'(0) = 1$ . No algebra opens $e^{\sin h} - 1$ ; recognizing the derivative is the way. (ii) Not yet of the shape $\frac{F(x) - F(c)}{x - c}$ : factor the constants first. $2\sin x - 1 = 2\left(\sin x - \frac{1}{2}\right)$ , $6x - \pi = 6\left(x - \frac{\pi}{6}\right)$ , and $\sin\frac{\pi}{6} = \frac{1}{2}$ . So the quotient is $\frac{2}{6} \cdot \frac{\sin x - \sin\frac{\pi}{6}}{x - \frac{\pi}{6}} \to \frac{1}{3}\cos\frac{\pi}{6} = \frac{\sqrt{3}}{6} \approx 0.2887$ .

c)

Add and subtract $f(1)$ in the numerator, then split: $\frac{f(1 + 2h) - f(1 - h)}{h} = 2 \cdot \frac{f(1 + 2h) - f(1)}{2h} + \frac{f(1 - h) - f(1)}{-h}$ . As $h \to 0$ , both $2h$ and $-h$ tend to $0$ , so each quotient tends to $f'(1)$ : the limit is $2f'(1) + f'(1) = 3f'(1) = -2$ . The increment must be the SAME in the numerator and the denominator of each quotient: that is why the factor $2$ and the sign are pulled out first.

Exercise 6: A curve with an exponential in it: tangents, symmetry, and a ceiling

6 answers to check

The curve $C$ of equation $x^2e^{y} + y = 1$ is shown in the figure, with the points $A(0, 1)$ , $B(1, 0)$ and $B'(-1, 0)$ .

a) Check that $A$ , $B$ and $B'$ lie on $C$ , and explain why $C$ is symmetric in the $y$ -axis. Find $\frac{dy}{dx}$ by implicit differentiation, then use the equation of $C$ to show that, for $x \ne 0$ , $\frac{dy}{dx} = -\frac{2(1 - y)}{x(2 - y)}$ , with no exponential left.
b) Find the tangent lines at $A$ , at $B$ and at $B'$ . Where does the tangent at $B$ meet the $y$ -axis?
c) Find $\frac{d^2y}{dx^2}$ at $A$ . Near $A$ , $C$ is the graph of a function $y(x)$ : does it have a local extremum at $x = 0$ ? Show that no point of $C$ lies above the line $y = 1$ .

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Answers

a) $\frac{dy}{dx} = -\frac{2xe^{y}}{x^2e^{y} + 1} = -\frac{2(1 - y)}{x(2 - y)}$ for $x \ne 0$
b) At $A$ : $y = 1$ ; at $B$ : $y = 1 - x$ , which passes through $A(0, 1)$ ; at $B'$ : $y = 1 + x$
c) $y'' = -2e \approx -5.4366$ at $A$ : a local maximum; $y = 1 - x^2e^{y} \le 1$ everywhere

a)

At $A$ : $0 + 1 = 1$ ; at $B$ : $1 \cdot e^{0} + 0 = 1$ ; at $B'$ : $(-1)^2e^{0} + 0 = 1$ . Replacing $x$ by $-x$ does not change $x^2$ , so $(x, y)$ is on $C$ exactly when $(-x, y)$ is: $C$ is symmetric in the $y$ -axis. Differentiate both sides with respect to $x$ , $y$ being a function of $x$ . The term $x^2e^{y}$ is a product, and $e^{y}$ a composition with inner function $y$ : $\frac{d}{dx}\left(x^2e^{y}\right) = 2xe^{y} + x^2e^{y}y'$ . So $2xe^{y} + x^2e^{y}y' + y' = 0$ ; collect and factor, $y'\left(x^2e^{y} + 1\right) = -2xe^{y}$ , and $\frac{dy}{dx} = -\frac{2xe^{y}}{x^2e^{y} + 1}$ .

Now the equation of $C$ does the algebra. On $C$ , $x^2e^{y} = 1 - y$ , so the denominator is $(1 - y) + 1 = 2 - y$ . For $x \ne 0$ , the same equation gives $e^{y} = \frac{1 - y}{x^2}$ , so the numerator is $2x \cdot \frac{1 - y}{x^2} = \frac{2(1 - y)}{x}$ . Hence $\frac{dy}{dx} = -\frac{2(1 - y)}{x(2 - y)}$ . The $x$ cancels once, not twice: $\frac{x}{x^2} = \frac{1}{x}$ .

b)

At $A$ : $\frac{dy}{dx} = -\frac{0}{0 + 1} = 0$ , horizontal tangent $y = 1$ . At $B$ : $\frac{dy}{dx} = -\frac{2(1 - 0)}{1(2 - 0)} = -1$ , tangent $y = -(x - 1) = 1 - x$ . At $B'$ : $-\frac{2}{(-1)(2)} = 1$ , tangent $y = 1 + x$ , the mirror image of the tangent at $B$ , as the symmetry requires: a reflection in the $y$ -axis changes the sign of every slope. The tangent at $B$ meets the $y$ -axis at $(0, 1)$ , which is the point $A$ itself: both tangents at $B$ and $B'$ pass through $A$ .

c)

Differentiate $2xe^{y} + x^2e^{y}y' + y' = 0$ once more. The first term gives $2e^{y} + 2xe^{y}y'$ ; the second, a product of three factors, gives $2xe^{y}y' + x^2e^{y}(y')^2 + x^2e^{y}y''$ ; the last gives $y''$ . At $A$ , where $x = 0$ , $y = 1$ and $y' = 0$ , everything vanishes except $2e + y'' = 0$ : $y'' = -2e \approx -5.4366$ . With $y'(0) = 0$ and $y''(0) < 0$ , the Second Derivative Test gives a local MAXIMUM of $y(x)$ at $x = 0$ . In fact the maximum is global: $y = 1 - x^2e^{y}$ and $x^2e^{y} \ge 0$ , so $y \le 1$ at every point of $C$ , with equality only at $x = 0$ . The curve never rises above its tangent at $A$ .

Exercise 7: The closed interval method with a cube root of a negative number

7 answers to check

Let $f(x) = x\sqrt[3]{x + 4} = x(x + 4)^{1/3}$ on the closed interval $[-5, 4]$ . A calculator is allowed, but several models answer ERROR to $(-1)^{1/3}$ : every value asked here is exact.

a) Write $f'(x)$ as one factored fraction and list the critical points of $f$ in $(-5, 4)$ , saying for each whether $f'$ is zero or undefined there.
b) Find the absolute maximum and minimum values of $f$ on $[-5, 4]$ , and where they occur.
c) Does $f$ have a local extremum at $x = -4$ ? Does $f$ have an absolute maximum on the OPEN interval $(-5, 4)$ ? Does this contradict the Extreme Value Theorem?

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Answers

a) $f'(x) = \frac{4(x + 3)}{3(x + 4)^{2/3}}$ ; critical points $-3$ ( $f' = 0$ ) and $-4$ ( $f'$ undefined)
b) $f(-5) = 5$ , $f(-4) = 0$ , $f(-3) = -3$ , $f(4) = 8$ : maximum $8$ at $x = 4$ , minimum $-3$ at $x = -3$
c) No extremum at $-4$ ( $f' < 0$ on both sides); no absolute maximum on $(-5, 4)$ ; no contradiction, the interval is not closed

a)

Product rule, and the chain rule on $(x + 4)^{1/3}$ (inner function $x + 4$ ): $f'(x) = (x + 4)^{1/3} + \frac{x}{3}(x + 4)^{-2/3}$ . Factor the LOWEST power, $(x + 4)^{-2/3}$ , and send it to the denominator: $f'(x) = \frac{(x + 4)^{-2/3}}{3}\left[3(x + 4) + x\right] = \frac{4x + 12}{3(x + 4)^{2/3}} = \frac{4(x + 3)}{3(x + 4)^{2/3}}$ . Critical points, interior points where $f'$ is zero or undefined: $x = -3$ , where the numerator vanishes, and $x = -4$ , where the denominator vanishes while $f(-4) = 0$ exists (a vertical tangent). Solving only $f'(x) = 0$ loses $-4$ .

b)

Closed interval method. $f$ is continuous on $[-5, 4]$ : a polynomial times a cube root, which is defined for every real number. Table of values of $f$ at the critical points and the endpoints: $f(-5) = -5\sqrt[3]{-1} = (-5)(-1) = 5$ , $f(-4) = 0$ , $f(-3) = -3\sqrt[3]{1} = -3$ , $f(4) = 4\sqrt[3]{8} = 8$ . The absolute maximum is $8$ , at $x = 4$ ; the absolute minimum is $-3$ , at $x = -3$ . A calculator that refuses $(-1)^{1/3}$ tempts one to drop the endpoint $-5$ : the cube root of a negative number exists, $\sqrt[3]{-1} = -1$ , and $f(-5) = 5$ is a real candidate, the second largest value.

c)

The denominator $3(x + 4)^{2/3}$ is positive on both sides of $-4$ , and the numerator $4(x + 3)$ is negative there: $f' < 0$ on $(-5, -4)$ and on $(-4, -3)$ . So $f$ keeps decreasing through $-4$ : no local extremum, only a vertical tangent. A critical point is a candidate, not a verdict. On the open interval $(-5, 4)$ , $f$ increases on $[-3, 4)$ and takes values as close to $8$ as we like, but never $8$ , since $x = 4$ is excluded; and $f(x) < 5 < 8$ on $(-5, -3]$ . So $f$ has NO absolute maximum there; the minimum $-3$ at $x = -3$ survives. No contradiction: the Extreme Value Theorem requires a CLOSED interval, and that hypothesis fails.

Exercise 8: A complete study: a root, a line, and two slant asymptotes

9 answers to check

Let $f(x) = \sqrt{x^2 + 4} - \dfrac{x}{2}$ . The figure shows its graph and two dashed lines. Every feature must be justified by a computation.

a) Give the domain and the $y$ -intercept of $f$ . Show that $y = \frac{x}{2}$ is a slant asymptote as $x \to \infty$ and $y = -\frac{3x}{2}$ one as $x \to -\infty$ , and on which side of each the graph lies.
b) Find $f'(x)$ and the critical number, checking every candidate that squaring produces. Give the intervals of increase and decrease and the absolute minimum value, exactly.
c) Find $f''(x)$ by factoring the lowest power, and the concavity. Does the graph have a point of inflection? Give the range of $f$ .
d) Solve $f(x) = 2$ exactly.

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Answers

a) Domain $\mathbb{R}$ , $f(0) = 2$ ; $f(x) - \frac{x}{2} = \frac{4}{\sqrt{x^2 + 4} + x} \to 0^+$ and $f(x) + \frac{3x}{2} = \frac{4}{\sqrt{x^2 + 4} - x} \to 0^+$ : the graph is above both
b) Critical number $\frac{2}{\sqrt{3}}$ ( $-\frac{2}{\sqrt{3}}$ rejected); decreasing on $\left(-\infty, \frac{2}{\sqrt{3}}\right]$ , increasing after; minimum $\sqrt{3}$
c) $f''(x) = \frac{4}{(x^2 + 4)^{3/2}} > 0$ : concave up everywhere, no inflection; range $[\sqrt{3}, \infty)$
d) $x = 0$ or $x = \frac{8}{3}$

a)

$x^2 + 4 > 0$ for every $x$ : the domain is $\mathbb{R}$ , $f$ is continuous, and there is no vertical asymptote. $f(0) = \sqrt{4} = 2$ . As $x \to \infty$ : $f(x) - \frac{x}{2} = \sqrt{x^2 + 4} - x$ , of the form $\infty - \infty$ . Conjugate: $\sqrt{x^2 + 4} - x = \frac{(x^2 + 4) - x^2}{\sqrt{x^2 + 4} + x} = \frac{4}{\sqrt{x^2 + 4} + x} \to 0$ , so $y = \frac{x}{2}$ is a slant asymptote on the right. As $x \to -\infty$ : $f(x) - \left(-\frac{3x}{2}\right) = \sqrt{x^2 + 4} + x = \frac{4}{\sqrt{x^2 + 4} - x} \to 0$ , the denominator tending to $+\infty$ because $-x > 0$ ; so $y = -\frac{3x}{2}$ is a slant asymptote on the left.

Two different slopes, because $\sqrt{x^2 + 4}$ behaves like $\sqrt{x^2} = |x|$ , which is $x$ on the right and $-x$ on the left: $|x| - \frac{x}{2}$ is $\frac{x}{2}$ on one side and $-\frac{3x}{2}$ on the other. Position: $\sqrt{x^2 + 4} > \sqrt{x^2} = |x|$ , so both differences $\sqrt{x^2 + 4} - x$ and $\sqrt{x^2 + 4} + x$ are positive for EVERY $x$ : the graph lies above both lines everywhere.

b)

$f'(x) = \frac{x}{\sqrt{x^2 + 4}} - \frac{1}{2} = \frac{2x - \sqrt{x^2 + 4}}{2\sqrt{x^2 + 4}}$ , defined everywhere. $f'(x) = 0$ means $2x = \sqrt{x^2 + 4}$ , which forces $x \ge 0$ , a root never being negative. Squaring: $4x^2 = x^2 + 4$ , $x^2 = \frac{4}{3}$ , $x = \pm\frac{2}{\sqrt{3}}$ . Check each: at $x = -\frac{2}{\sqrt{3}}$ , $2x < 0 < \sqrt{x^2 + 4}$ , so it is NOT a solution: squaring created it, rejected. The only critical number is $\frac{2}{\sqrt{3}} = \frac{2\sqrt{3}}{3} \approx 1.1547$ .

Sign of the numerator $2x - \sqrt{x^2 + 4}$ : for $x \le 0$ it is negative; for $x > 0$ both terms are positive and it has the sign of the difference of their squares, $4x^2 - (x^2 + 4) = 3x^2 - 4$ . So $f' < 0$ on $\left(-\infty, \frac{2}{\sqrt{3}}\right)$ and $f' > 0$ on $\left(\frac{2}{\sqrt{3}}, \infty\right)$ : $f$ decreases, then increases. The minimum is ABSOLUTE, since $f$ decreases on everything to its left and increases on everything to its right. Its value: $f\left(\frac{2}{\sqrt{3}}\right) = \sqrt{\frac{4}{3} + 4} - \frac{1}{\sqrt{3}} = \frac{4}{\sqrt{3}} - \frac{1}{\sqrt{3}} = \frac{3}{\sqrt{3}} = \sqrt{3}$ .

c)

Write $f'(x) = x(x^2 + 4)^{-1/2} - \frac{1}{2}$ and use the product rule: $f''(x) = (x^2 + 4)^{-1/2} - x^2(x^2 + 4)^{-3/2}$ . Factor the lowest power $(x^2 + 4)^{-3/2}$ : $f''(x) = (x^2 + 4)^{-3/2}\left[(x^2 + 4) - x^2\right] = \frac{4}{(x^2 + 4)^{3/2}} > 0$ . The graph is concave up on $\mathbb{R}$ and has NO point of inflection. With b), the range is $[\sqrt{3}, \infty)$ : the minimum is reached, and $f(x) \to \infty$ at both ends, the graph staying above two lines that rise.

d)

$\sqrt{x^2 + 4} = 2 + \frac{x}{2}$ requires $2 + \frac{x}{2} \ge 0$ , that is $x \ge -4$ . Squaring: $x^2 + 4 = 4 + 2x + \frac{x^2}{4}$ , so $\frac{3x^2}{4} - 2x = 0$ and $x\left(\frac{3x}{4} - 2\right) = 0$ : $x = 0$ or $x = \frac{8}{3}$ . Both satisfy $x \ge -4$ , and both check: $f(0) = 2$ and $f\left(\frac{8}{3}\right) = \sqrt{\frac{64}{9} + 4} - \frac{4}{3} = \frac{10}{3} - \frac{4}{3} = 2$ . Two solutions, one on each side of the minimum, as $2 > \sqrt{3}$ predicts. Dividing $\frac{3x^2}{4} = 2x$ by $x$ loses the solution $0$ .

Exercise 9: Related rates on a clock face: the tips of the hands

9 answers to check

The minute hand of a wall clock is $6$ cm long and its hour hand $4$ cm, both measured from the centre $O$ of the dial. Let $\theta$ be the angle between the hands, in radians, and $D$ the distance between their tips, in cm, both functions of the time $t$ in hours. The figure shows the hands a few minutes after 3:00. Calculator allowed; give decimals to two places.

a) Give the angular speed of each hand in rad/h, and explain why $\frac{d\theta}{dt} = -\frac{11\pi}{6}$ rad/h from 3:00 until the hands overlap. Use the law of cosines to express $D^2$ in terms of $\theta$ , and give $D$ at 3:00.
b) How fast is the distance between the tips changing at 3:00?
c) How fast is it changing at 4:00, when $\theta = \frac{2\pi}{3}$ and the angle is still closing?
d) How many minutes after 3:00 do the hands first overlap? Give $D$ and $\frac{dD}{dt}$ at that instant.
e) Let $A$ be the area of the triangle formed by $O$ and the two tips. How fast is $A$ changing at 3:00, and at 4:00?

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Answers

a) $2\pi$ and $\frac{\pi}{6}$ rad/h; $D^2 = 52 - 48\cos\theta$ ; $D = 2\sqrt{13} \approx 7.21$ cm at 3:00
b) $\frac{dD}{dt} = -\frac{22\pi}{\sqrt{13}} \approx -19.17$ cm/h: the tips get closer
c) $\frac{dD}{dt} = -\frac{11\sqrt{3}\,\pi}{\sqrt{19}} \approx -13.73$ cm/h
d) $\frac{180}{11} \approx 16.36$ min after 3:00; $D = 2$ cm, $\frac{dD}{dt} = 0$
e) $\frac{dA}{dt} = 0$ at 3:00; $11\pi \approx 34.56$ cm $^2$ /h at 4:00

a)

The minute hand turns once, $2\pi$ rad, per hour; the hour hand once per $12$ hours, $\frac{2\pi}{12} = \frac{\pi}{6}$ rad/h. Both turn clockwise. After 3:00 the minute hand, behind the hour hand, closes the gap at the DIFFERENCE of the two speeds, so $\theta$ decreases: $\frac{d\theta}{dt} = \frac{\pi}{6} - 2\pi = -\frac{11\pi}{6}$ rad/h, a negative rate. In the triangle with sides $6$ and $4$ around the angle $\theta$ , the law of cosines gives, at every instant, $D^2 = 6^2 + 4^2 - 2(6)(4)\cos\theta = 52 - 48\cos\theta$ . At 3:00, $\theta = \frac{\pi}{2}$ and $\cos\theta = 0$ : $D = \sqrt{52} = 2\sqrt{13} \approx 7.21$ cm.

b)

Differentiate the relation with respect to $t$ , the chain rule on both sides: $2D\frac{dD}{dt} = 48\sin\theta\,\frac{d\theta}{dt}$ , so $\frac{dD}{dt} = \frac{24\sin\theta}{D}\,\frac{d\theta}{dt}$ . Only now substitute the instant: at 3:00, $\frac{dD}{dt} = \frac{24 \cdot 1}{2\sqrt{13}} \cdot \left(-\frac{11\pi}{6}\right) = -\frac{22\pi}{\sqrt{13}} \approx -19.17$ cm/h. The tips are getting closer, at about $19$ cm per hour. With $\frac{d\theta}{dt}$ in degrees per hour the answer would be $57$ times too large: $\frac{d}{d\theta}\cos\theta = -\sin\theta$ holds in radians only.

c)

At 4:00, $\cos\frac{2\pi}{3} = -\frac{1}{2}$ and $\sin\frac{2\pi}{3} = \frac{\sqrt{3}}{2}$ : $D^2 = 52 + 24 = 76$ , $D = 2\sqrt{19} \approx 8.72$ cm. Then $\frac{dD}{dt} = \frac{24 \cdot \frac{\sqrt{3}}{2}}{2\sqrt{19}} \cdot \left(-\frac{11\pi}{6}\right) = -\frac{11\sqrt{3}\,\pi}{\sqrt{19}} \approx -13.73$ cm/h. The angle closes at the same rate as at 3:00, yet the distance shrinks more slowly: the factor $\frac{24\sin\theta}{D}$ is smaller, $\sin\theta$ being smaller and $D$ larger.

d)

Measure positions clockwise from 12, in radians, $t$ hours after 3:00: the minute hand is at $2\pi t$ , the hour hand at $\frac{\pi}{2} + \frac{\pi}{6}t$ . They overlap when $2\pi t = \frac{\pi}{2} + \frac{\pi}{6}t$ , that is $\frac{11\pi}{6}t = \frac{\pi}{2}$ , $t = \frac{3}{11}$ h $= \frac{180}{11} \approx 16.36$ min, at about 3:16:22. Then $\theta = 0$ , $D^2 = 52 - 48 = 4$ , $D = 2$ cm, the difference of the two lengths, and $\frac{dD}{dt} = \frac{24\sin 0}{2}\cdot\left(-\frac{11\pi}{6}\right) = 0$ : the distance reaches its minimum and stops decreasing.

e)

$A = \frac{1}{2}(6)(4)\sin\theta = 12\sin\theta$ , so $\frac{dA}{dt} = 12\cos\theta\,\frac{d\theta}{dt}$ . At 3:00, $\cos\frac{\pi}{2} = 0$ : $\frac{dA}{dt} = 0$ , the area is momentarily stationary, at its largest value $12$ cm $^2$ with the hands at a right angle. At 4:00: $\frac{dA}{dt} = 12\left(-\frac{1}{2}\right)\left(-\frac{11\pi}{6}\right) = 11\pi \approx 34.56$ cm $^2$ /h. Two negative factors give a positive rate: the area GROWS while $\theta$ closes from $\frac{2\pi}{3}$ toward $\frac{\pi}{2}$ .

Exercise 10: Linearization and error: the power of a wind turbine

8 answers to check

The power delivered by a small wind turbine is modelled by $P(v) = 0.5v^3$ kilowatts when the wind blows at $v$ metres per second, for $4 \le v \le 14$ . The figure shows $P$ and its tangent at $v = 10$ . Calculator allowed; give values to two decimals.

a) Find the linearization $L(v)$ of $P$ at $v = 10$ . Use it to estimate the power at $v = 10.4$ m/s and at $v = 9.6$ m/s, compare with the exact values, and explain the side of both errors with $P''$ .
b) The anemometer reads $v = 10$ m/s with a possible error of $0.2$ m/s. Use a differential to estimate the largest error in the computed power. Show that $\frac{dP}{P} = 3\frac{dv}{v}$ and give the percentage error.
c) The operator needs the wind speed that gives a required power. Solve $P = 0.5v^3$ for $v$ , linearize $v(P)$ at $P = 500$ kW, and estimate the wind speed that gives $530$ kW. Compare with the calculator. How is the slope of this linearization related to $P'(10)$ ?
d) How precisely must the wind speed be measured at $10$ m/s for the power to be known within $1.5\%$ ?

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Answers

a) $L(v) = 500 + 150(v - 10)$ ; $560$ and $440$ kW against $562.43$ and $442.37$ kW: both too small, $P'' = 3v > 0$
b) $dP = 30$ kW; $\frac{dP}{P} = 3 \cdot 2\% = 6\%$
c) $v = (2P)^{1/3}$ , slope $\frac{1}{150} = \frac{1}{P'(10)}$ ; estimate $10.2$ m/s against $10.20$ m/s ( $1060^{1/3} \approx 10.196$ )
d) $\frac{dv}{v} = 0.5\%$ : within $0.05$ m/s

a)

$P(10) = 0.5 \cdot 1000 = 500$ kW and $P'(v) = 1.5v^2$ , so $P'(10) = 150$ kW per m/s: $L(v) = 500 + 150(v - 10)$ . At $10.4$ : $L = 500 + 150(0.4) = 560$ kW, while $P(10.4) = 0.5 \cdot 10.4^3 = 562.432$ , about $562.43$ kW. At $9.6$ : $L = 500 - 60 = 440$ kW, while $P(9.6) = 0.5 \cdot 884.736 = 442.368$ , about $442.37$ kW. Both estimates are too small: $P''(v) = 3v > 0$ , so $P$ is concave up and its tangent lies BELOW the curve on both sides of $10$ , as the figure shows. The sign of $v - 10$ does not decide the side; the concavity does.

b)

$dP = P'(10)\,dv = 150 \cdot 0.2 = 30$ kW: the computed $500$ kW may be off by about $30$ kW. In relative terms, $\frac{dP}{P} = \frac{1.5v^2\,dv}{0.5v^3} = 3\frac{dv}{v}$ : the constant $0.5$ cancels and the exponent $3$ comes down as a factor. Here $\frac{dv}{v} = \frac{0.2}{10} = 2\%$ , so $\frac{dP}{P} = 6\%$ . A power of $3$ triples a relative error: the anemometer is three times more precise than the power it predicts.

c)

$v^3 = 2P$ , so $v(P) = (2P)^{1/3}$ . Chain rule, inner function $2P$ : $v'(P) = \frac{1}{3}(2P)^{-2/3} \cdot 2 = \frac{2}{3}(2P)^{-2/3}$ . At $P = 500$ : $(1000)^{-2/3} = \frac{1}{\left(\sqrt[3]{1000}\right)^2} = \frac{1}{100}$ , so $v'(500) = \frac{2}{300} = \frac{1}{150}$ m/s per kW. The negative fractional exponent is read in two steps, the cube root first, then the square and the reciprocal. The linearization is $v \approx 10 + \frac{P - 500}{150}$ , and for $530$ kW: $v \approx 10 + \frac{30}{150} = 10.2$ m/s. The calculator gives $1060^{1/3} \approx 10.196$ m/s: this time the estimate is slightly too LARGE, because $v(P)$ , a cube root, is concave down.

The slope $\frac{1}{150}$ is exactly $\frac{1}{P'(10)}$ : $v(P)$ is the inverse function of $P(v)$ , and the derivative of an inverse at $b = P(a)$ is $\frac{1}{P'(a)}$ . Graphically, the graph of $v(P)$ is the graph of $P(v)$ with the two axes exchanged, and exchanging the axes inverts every slope: $150$ kW per m/s becomes $\frac{1}{150}$ m/s per kW.

d)

We need $\left|\frac{dP}{P}\right| = 3\left|\frac{dv}{v}\right| \le 1.5\%$ , that is $\left|\frac{dv}{v}\right| \le 0.5\%$ . At $10$ m/s: $|dv| \le 0.005 \cdot 10 = 0.05$ m/s. Reading the relation in this direction is the whole point of the exponent: to gain a factor on the power, the wind must be measured three times better.

Exercise 11: Applied optimization: where to cut the wire, and where not to

7 answers to check

A wire $60$ cm long is cut into two pieces. The piece of length $x$ is bent into a square, the other, of length $60 - x$ , into an equilateral triangle, as in the figure. The values $x = 0$ (all triangle) and $x = 60$ (all square) are allowed: the wire is then not cut at all. Give lengths and areas to two decimals.

a) Show that the total area enclosed is $A(x) = \dfrac{x^2}{16} + \dfrac{\sqrt{3}}{36}(60 - x)^2$ cm $^2$ , give its domain, and compute $A(0)$ .
b) Find the critical number of $A$ exactly, with a rationalized denominator, then to two decimals.
c) Where should the wire be cut to make the total area as small as possible? As large as possible? Give both areas, and justify that they are absolute.
d) Now the second piece is bent into a CIRCLE instead of a triangle. Find where to cut for the smallest total area, and show that the largest total area is obtained without cutting, with the circle alone. Rank the three shapes by the area one length of wire encloses.

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Answers

a) Square of side $\frac{x}{4}$ , triangle of side $\frac{60 - x}{3}$ ; domain $[0, 60]$ ; $A(0) = 100\sqrt{3} \approx 173.21$ cm $^2$
b) $x = \frac{240\sqrt{3}}{9 + 4\sqrt{3}} = \frac{720\sqrt{3} - 960}{11} \approx 26.10$ cm
c) Minimum $\approx 97.87$ cm $^2$ at $x \approx 26.10$ ; maximum $225$ cm $^2$ at $x = 60$ : no cut, all square
d) Minimum at $x = \frac{240}{\pi + 4} \approx 33.61$ cm; maximum $\frac{900}{\pi} \approx 286.48$ cm $^2$ at $x = 0$ , all circle; circle, then square, then triangle

a)

The square has perimeter $x$ , so its side is $\frac{x}{4}$ and its area $\frac{x^2}{16}$ . The triangle has perimeter $60 - x$ , so its side is $s = \frac{60 - x}{3}$ , and an equilateral triangle of side $s$ has height $\frac{\sqrt{3}}{2}s$ and area $\frac{\sqrt{3}}{4}s^2 = \frac{\sqrt{3}}{4} \cdot \frac{(60 - x)^2}{9} = \frac{\sqrt{3}}{36}(60 - x)^2$ . The domain is the closed interval $[0, 60]$ , both ends included by the statement, and $A$ is continuous there. $A(0) = \frac{\sqrt{3}}{36} \cdot 3600 = 100\sqrt{3} \approx 173.21$ cm $^2$ .

b)

Chain rule on $(60 - x)^2$ , inner function $60 - x$ of derivative $-1$ : $A'(x) = \frac{x}{8} - \frac{\sqrt{3}}{18}(60 - x)$ . Set it to $0$ and collect the $x$ terms: $x\left(\frac{1}{8} + \frac{\sqrt{3}}{18}\right) = \frac{60\sqrt{3}}{18}$ . Multiply both sides by $72$ , the common denominator, to clear the fractions: $x(9 + 4\sqrt{3}) = 240\sqrt{3}$ , so $x = \frac{240\sqrt{3}}{9 + 4\sqrt{3}}$ . Rationalize with the conjugate $9 - 4\sqrt{3}$ : the denominator becomes $81 - 48 = 33$ , and $x = \frac{240\sqrt{3}(9 - 4\sqrt{3})}{33} = \frac{2160\sqrt{3} - 2880}{33} = \frac{720\sqrt{3} - 960}{11} \approx 26.10$ cm. It lies in $(0, 60)$ .

c)

Closed Interval Method, $A$ being continuous on $[0, 60]$ : $A(0) \approx 173.21$ , $A(26.10) \approx 97.87$ and $A(60) = \frac{3600}{16} = 225$ cm $^2$ . The smallest area, about $97.87$ cm $^2$ , comes from cutting at $x \approx 26.10$ cm (the exact value is $\frac{900\sqrt{3}}{9 + 4\sqrt{3}}$ ). The largest, $225$ cm $^2$ , comes from NOT cutting: all the wire into the square. Also, $A''(x) = \frac{1}{8} + \frac{\sqrt{3}}{18} > 0$ : $A$ is concave up, so its only critical number is a minimum and the maximum must be at an endpoint. Stopping at $A'(x) = 0$ gives the cut that encloses the LEAST area, the opposite of what one wanted.

d)

With a circle of circumference $60 - x$ , the radius is $\frac{60 - x}{2\pi}$ and the area $\pi\left(\frac{60 - x}{2\pi}\right)^2 = \frac{(60 - x)^2}{4\pi}$ . So $A(x) = \frac{x^2}{16} + \frac{(60 - x)^2}{4\pi}$ on $[0, 60]$ , and $A'(x) = \frac{x}{8} - \frac{60 - x}{2\pi} = 0$ gives $2\pi x = 8(60 - x)$ , $x = \frac{480}{2\pi + 8} = \frac{240}{\pi + 4} \approx 33.61$ cm, the cut for the smallest area. Endpoints: $A(0) = \frac{3600}{4\pi} = \frac{900}{\pi} \approx 286.48$ cm $^2$ and $A(60) = 225$ cm $^2$ . The largest area is $\frac{900}{\pi}$ , with no cut and the circle alone. For $60$ cm of wire: circle $286.48$ , square $225$ , equilateral triangle $173.21$ cm $^2$ . The rounder the shape, the more it encloses.

Exercise 12: L'Hôpital's Rule in economics: from a CES function to Cobb-Douglas

8 answers to check

A firm uses $K = 16$ units of capital and $L = 81$ units of labour. Economists model its output with a CES production function (constant elasticity of substitution): $Q(\rho) = \left(0.25 \cdot 16^{\rho} + 0.75 \cdot 81^{\rho}\right)^{1/\rho}$ , where the parameter $\rho \ne 0$ measures how easily capital and labour replace each other. The figure shows $Q$ as a function of $\rho$ , with a hole at $\rho = 0$ . Calculator allowed; decimals to two places.

a) Compute $Q(1)$ , $Q(-1)$ and $Q(2)$ .
b) Name the form of $Q(\rho)$ as $\rho \to 0$ . Find $\lim_{\rho\to 0} Q(\rho)$ with L'Hôpital's Rule, and show that it equals $16^{1/4} \cdot 81^{3/4}$ , the output of the Cobb-Douglas function $K^{1/4}L^{3/4}$ .
c) Find $\lim_{\rho\to\infty} Q(\rho)$ and $\lim_{\rho\to -\infty} Q(\rho)$ , naming each form first and then factoring out the dominant power. Interpret the second limit.
d) Prove that for all $K > 0$ , $L > 0$ and $0 < a < 1$ , $\displaystyle\lim_{\rho\to 0}\left(aK^{\rho} + (1 - a)L^{\rho}\right)^{1/\rho} = K^aL^{1 - a}$ .
e) Compute $Q(0.01)$ and $Q(-0.01)$ , and say how the hole of the figure should be filled.

Show the solution

Answers

a) $Q(1) = 64.75$ , $Q(-1) \approx 40.19$ , $Q(2) \approx 70.60$
b) $1^{\pm\infty}$ ; $\ln Q \to 0.25\ln 16 + 0.75\ln 81 = \ln 54$ : the limit is $54 = 16^{1/4} \cdot 81^{3/4}$
c) $\infty^0$ at $+\infty$ : limit $81$ ; $0^0$ at $-\infty$ : limit $16$ , the scarcer input
d) $\ln$ of the expression $\to a\ln K + (1 - a)\ln L = \ln\left(K^aL^{1 - a}\right)$
e) $Q(0.01) \approx 54.13$ , $Q(-0.01) \approx 53.87$ ; define $Q(0) = 54$ , the continuous extension

a)

$Q(1) = 0.25 \cdot 16 + 0.75 \cdot 81 = 4 + 60.75 = 64.75$ , the weighted average of the inputs. $Q(-1) = \left(\frac{0.25}{16} + \frac{0.75}{81}\right)^{-1} = \frac{1}{0.015625 + 0.009259\ldots} \approx 40.19$ : the exponent $-1$ inverts the WHOLE sum, not each term. $Q(2) = \sqrt{0.25 \cdot 256 + 0.75 \cdot 6561} = \sqrt{4984.75} \approx 70.60$ .

b)

As $\rho \to 0$ the base tends to $0.25 + 0.75 = 1$ and the exponent $\frac{1}{\rho}$ to $\pm\infty$ : the form $1^{\pm\infty}$ , indeterminate, and the answer is NOT $1$ . The base is positive, so take the logarithm: $\ln Q(\rho) = \frac{\ln\left(0.25 \cdot 16^{\rho} + 0.75 \cdot 81^{\rho}\right)}{\rho}$ , of the form $\frac{\ln 1}{0} = \frac{0}{0}$ . L'Hôpital's Rule, the variable being $\rho$ and the bases constant, with $\frac{d}{d\rho}a^{\rho} = a^{\rho}\ln a$ : the derivative of the numerator is $\frac{0.25 \cdot 16^{\rho}\ln 16 + 0.75 \cdot 81^{\rho}\ln 81}{0.25 \cdot 16^{\rho} + 0.75 \cdot 81^{\rho}}$ , that of the denominator is $1$ .

This quotient is continuous at $\rho = 0$ , where it equals $0.25\ln 16 + 0.75\ln 81$ . Laws of logarithms: $0.25\ln 16 = \ln 16^{1/4} = \ln 2$ and $0.75\ln 81 = \ln 81^{3/4} = \ln 27$ , so $\ln Q(\rho) \to \ln 2 + \ln 27 = \ln 54$ . The exponential being continuous, $Q(\rho) = e^{\ln Q(\rho)} \to 54 = 16^{1/4} \cdot 81^{3/4}$ , the Cobb-Douglas output.

c)

As $\rho \to \infty$ the base tends to $\infty$ and the exponent to $0$ : the form $\infty^0$ . Factor the dominant power $81^{\rho}$ out of the base, using $(ab)^{1/\rho} = a^{1/\rho}b^{1/\rho}$ : $Q(\rho) = \left(81^{\rho}\right)^{1/\rho}\left(0.75 + 0.25\left(\frac{16}{81}\right)^{\rho}\right)^{1/\rho} = 81\left(0.75 + 0.25\left(\frac{16}{81}\right)^{\rho}\right)^{1/\rho}$ . Now $\left(\frac{16}{81}\right)^{\rho} \to 0$ , the base tends to $0.75$ and the exponent to $0$ : $0.75^0 = 1$ is a DETERMINATE form, and the limit is $81$ .

As $\rho \to -\infty$ , $16^{\rho}$ and $81^{\rho}$ both tend to $0$ : the base tends to $0^+$ and the exponent to $0^-$ , the form $0^0$ . The dominant term is now $16^{\rho} = \frac{1}{16^{|\rho|}}$ , which vanishes more slowly than $81^{\rho}$ : $Q(\rho) = 16\left(0.25 + 0.75\left(\frac{81}{16}\right)^{\rho}\right)^{1/\rho} \to 16 \cdot 0.25^0 = 16$ . When capital and labour cannot replace each other at all, output is limited by the scarcer input, the $16$ units of capital. As $\rho$ runs over $\mathbb{R}$ , $Q$ goes from $16$ to $81$ , as the figure shows.

d)

Same three steps. The base $aK^{\rho} + (1 - a)L^{\rho}$ tends to $a + (1 - a) = 1$ and the exponent to $\pm\infty$ : the form $1^{\pm\infty}$ . Its logarithm, $\frac{\ln\left(aK^{\rho} + (1 - a)L^{\rho}\right)}{\rho}$ , has the form $\frac{0}{0}$ , and one round of the rule gives $\frac{aK^{\rho}\ln K + (1 - a)L^{\rho}\ln L}{aK^{\rho} + (1 - a)L^{\rho}} \to a\ln K + (1 - a)\ln L = \ln\left(K^aL^{1 - a}\right)$ . The exponential, continuous, gives the limit $K^aL^{1 - a}$ .

e)

$Q(0.01) \approx 54.13$ and $Q(-0.01) \approx 53.87$ , one on each side of $54$ : the values close in on the limit of b). $Q$ is not defined at $0$ , but the limit exists and is finite, so setting $Q(0) = 54$ gives the continuous extension of $Q$ : the hole is filled by the Cobb-Douglas output. Typing $\rho = 0$ into the formula gives $1^{1/0}$ , an error, which is why the limit is needed at all.

Practice final with full solutions (MATH 203)

Part A: techniques (/20)

Exercise 1: Four limits, and the algebra that opens each one

Exercise 2: Three derivatives: every rule named, every result simplified

Exercise 3: A moving exponent, and the slope of an inverse with no formula

Part B: theorems and reasoning (/32)

Exercise 4: Two seams to close, then a root located by bisection

Exercise 5: The derivative from its definition, then read backwards

Exercise 6: A curve with an exponential in it: tangents, symmetry, and a ceiling

Exercise 7: The closed interval method with a cube root of a negative number

Exercise 8: A complete study: a root, a line, and two slant asymptotes

Part C: applications (/48)

Exercise 9: Related rates on a clock face: the tips of the hands

Exercise 10: Linearization and error: the power of a wind turbine

Exercise 11: Applied optimization: where to cut the wire, and where not to

Exercise 12: L'Hôpital's Rule in economics: from a CES function to Cobb-Douglas

See also

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