Exercise 1: From the slope of a secant to the slope of the tangent

The tangent line to a curve at a point $P$ is the limiting position of the secant lines $PQ$ as $Q$ slides along the curve toward $P$ . Its slope is therefore a LIMIT of slopes of secants: if $P = (a, f(a))$ and $Q = (a + h, f(a + h))$ , then $m = \lim_{h\to 0} \frac{f(a+h) - f(a)}{h}$ , provided the limit exists.

The figure shows $f(x) = x^2 - 4x + 5$ , the point $P(3, 2)$ and three secants: $Q_1$ and $Q_2$ to the right of $P$ , $Q_3$ to its left. No calculator is needed: every slope is a quotient of small integers or decimals.

a) Compute the slope of the secant $PQ$ when $Q$ has $x$ -coordinate $4.5$ , then $4$ , then $3.5$ . Check the first two on the figure.
b) Same question when $Q$ has $x$ -coordinate $2$ , then $2.5$ . What do the two lists of slopes suggest?
c) Write the slope of the secant for a general $h \ne 0$ , simplify it, and find the slope $m$ of the tangent at $P$ by letting $h \to 0$ .
d) Write the equation of the tangent at $P$ . Show that the curve lies above this line and touches it only at $P$ .
e) A student writes: slope at $P$ $= \frac{f(3 + 0) - f(3)}{0} = \frac{0}{0}$ , so the tangent at $P$ has no slope. Explain what is wrong.

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Answers

a) $3.5$ , $3$ and $2.5$
b) $1$ and $1.5$ : the slopes close in on $2$ from both sides.
c) $\frac{f(3+h) - f(3)}{h} = 2 + h$ for $h \ne 0$ , so $m = 2$
d) $y = 2x - 4$ ; $f(x) - (2x - 4) = (x - 3)^2 \ge 0$ , zero only at $x = 3$
e) $h = 0$ is excluded by the definition: $\frac{0}{0}$ is the form of every tangent slope, the limit of $2 + h$ is $2$ .

a) $f(3) = 9 - 12 + 5 = 2$ . For $Q$ at $x = 4.5$ : $f(4.5) = 20.25 - 18 + 5 = 7.25$ , so the slope is $\frac{7.25 - 2}{4.5 - 3} = \frac{5.25}{1.5} = 3.5$ . At $x = 4$ : $f(4) = 5$ , slope $\frac{5 - 2}{1} = 3$ . At $x = 3.5$ : $f(3.5) = 12.25 - 14 + 5 = 3.25$ , slope $\frac{1.25}{0.5} = 2.5$ . On the figure, from $P(3, 2)$ to $Q_2(4, 5)$ the secant rises $3$ grid units for $1$ across, and from $P$ to $Q_1$ it rises a little over $5$ for $1.5$ across. As $Q$ comes closer to $P$ from the right, the secants turn clockwise and their slope decreases.

b) At $x = 2$ : $f(2) = 1$ , slope $\frac{1 - 2}{2 - 3} = \frac{-1}{-1} = 1$ ; the secant $PQ_3$ of the figure. At $x = 2.5$ : $f(2.5) = 6.25 - 10 + 5 = 1.25$ , slope $\frac{1.25 - 2}{-0.5} = 1.5$ . The slopes from the right, $3.5, 3, 2.5$ , decrease; the slopes from the left, $1, 1.5$ , increase; both lists move toward $2$ . A table of secant slopes SUGGESTS the answer, it never proves it: that is the job of the limit in c). Note the two signs in the quotient on the left: $\Delta y$ and $\Delta x$ are both negative, and forgetting one of them gives $-1$ instead of $1$ .

c) For $h \ne 0$ : $f(3 + h) = (3 + h)^2 - 4(3 + h) + 5 = 9 + 6h + h^2 - 12 - 4h + 5 = 2 + 2h + h^2$ . So $\frac{f(3+h) - f(3)}{h} = \frac{2h + h^2}{h} = 2 + h$ , the cancellation being legitimate precisely because $h \ne 0$ . Then $m = \lim_{h\to 0} (2 + h) = 2$ . Check against a) and b): the quotient $2 + h$ gives $3.5$ for $h = 1.5$ , $3$ for $h = 1$ , $1$ for $h = -1$ , exactly the slopes computed one by one. The whole chapter is in these three lines: expand $f(a + h)$ , cancel $h$ , THEN let $h \to 0$ .

d) The tangent passes through $P(3, 2)$ with slope $2$ : $y - 2 = 2(x - 3)$ , that is $y = 2x - 4$ . The vertical gap between curve and line is $f(x) - (2x - 4) = x^2 - 6x + 9 = (x - 3)^2$ , which is $\ge 0$ for every $x$ and equals $0$ only at $x = 3$ . So the parabola lies above its tangent and touches it once. For a parabola this is always the picture; Exercise 8 shows it is NOT a property of tangent lines in general.

e) The definition never sets $h = 0$ : the quotient $\frac{f(3 + h) - f(3)}{h}$ is the slope of a secant, and there is no secant when $Q = P$ . Substituting $h = 0$ produces $\frac{0}{0}$ for EVERY function and EVERY point, since the numerator $f(a) - f(a)$ is always $0$ : this form says nothing, and it is the reason why a derivative is a limit and not a value. The student had to simplify first, $\frac{2h + h^2}{h} = 2 + h$ for $h \ne 0$ , and only then let $h \to 0$ . Writing $\frac{0}{0}$ as a conclusion costs the whole part on a MATH 140 paper.

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Exercise 2: The derivative at a point, by the definition, in both forms

The derivative of $f$ at $a$ has two equivalent forms: $f'(a) = \lim_{h\to 0} \frac{f(a+h) - f(a)}{h}$ and $f'(a) = \lim_{x\to a} \frac{f(x) - f(a)}{x - a}$ , the second obtained from the first with $x = a + h$ . Both are limits of the form $\frac{0}{0}$ , and the whole work is to REWRITE the quotient until the factor that vanishes cancels.

The rewriting depends on the form of $f$ : expand a polynomial, factor out $(x - a)$ , multiply a root by its conjugate, put a fraction over a common denominator. Name the gesture on your paper. The differentiation rules of the next chapter are NOT allowed here, except to check an answer.

a) $f(x) = 2x^2 - 3x$ : find $f'(-1)$ with the $h$ form.
b) Find $f'(-1)$ again, with the $x \to a$ form, by factoring.
c) $g(x) = \sqrt{2x - 1}$ : find $g'(5)$ .
d) $k(x) = \frac{3}{x + 1}$ : find $k'(2)$ with the $x \to a$ form.
e) $m(x) = x^3 - x$ : find $m'(2)$ with the $x \to a$ form, by dividing $x^3 - x - 6$ by $x - 2$ .

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Answers

a) $\frac{f(-1+h) - f(-1)}{h} = -7 + 2h$ , so $f'(-1) = -7$
b) $\frac{2x^2 - 3x - 5}{x + 1} = 2x - 5$ for $x \ne -1$ , so $f'(-1) = -7$
c) $g'(5) = \lim_{h\to 0} \frac{2}{\sqrt{9 + 2h} + 3} = \frac{1}{3}$
d) $k'(2) = \lim_{x\to 2} \frac{-1}{x + 1} = -\frac{1}{3}$
e) $x^3 - x - 6 = (x - 2)(x^2 + 2x + 3)$ , so $m'(2) = 11$

a) $f(-1) = 2 + 3 = 5$ . Expand: $f(-1 + h) = 2(1 - 2h + h^2) - 3(-1 + h) = 2 - 4h + 2h^2 + 3 - 3h = 5 - 7h + 2h^2$ . So $\frac{f(-1+h) - f(-1)}{h} = \frac{-7h + 2h^2}{h} = -7 + 2h$ for $h \ne 0$ , and $f'(-1) = \lim_{h\to 0}(-7 + 2h) = -7$ . The classic slip is $f(-1 + h) = 2(-1 + h)^2 - 3(-1) + h$ : the $h$ must be substituted in EVERY occurrence of $x$ , with brackets. The constant terms always cancel ( $5 - 5$ ); if they do not, the expansion is wrong.

b) $\frac{f(x) - f(-1)}{x - (-1)} = \frac{2x^2 - 3x - 5}{x + 1}$ . Since $x = -1$ is a root of the numerator ( $2 + 3 - 5 = 0$ ), the factor $x + 1$ divides it: $2x^2 - 3x - 5 = (x + 1)(2x - 5)$ . So the quotient equals $2x - 5$ for $x \ne -1$ , and $f'(-1) = \lim_{x\to -1}(2x - 5) = -7$ , as in a). The root test is the check of the method: in the $x \to a$ form the numerator ALWAYS vanishes at $x = a$ , so $(x - a)$ ALWAYS divides a polynomial numerator. Check with the power rule, allowed as a check only: $f'(x) = 4x - 3$ gives $-7$ .

c) $g(5) = \sqrt 9 = 3$ and $g(5 + h) = \sqrt{9 + 2h}$ . The quotient $\frac{\sqrt{9 + 2h} - 3}{h}$ is of the form $\frac{0}{0}$ with a root: multiply by the conjugate. $\frac{(\sqrt{9 + 2h} - 3)(\sqrt{9 + 2h} + 3)}{h(\sqrt{9 + 2h} + 3)} = \frac{9 + 2h - 9}{h(\sqrt{9 + 2h} + 3)} = \frac{2}{\sqrt{9 + 2h} + 3}$ for $h \ne 0$ . The limit is $\frac{2}{3 + 3} = \frac{1}{3}$ . Never expand the conjugate in the denominator: it is the factor that stays, and it becomes $2\sqrt{9} = 6$ at the limit. The $2$ on top comes from the $2x$ inside the root; answering $\frac{1}{6}$ forgets it.

d) $k(2) = 1$ . $\frac{k(x) - k(2)}{x - 2} = \frac{\frac{3}{x+1} - 1}{x - 2} = \frac{\frac{3 - (x + 1)}{x + 1}}{x - 2} = \frac{2 - x}{(x + 1)(x - 2)}$ . Since $2 - x = -(x - 2)$ , this is $\frac{-1}{x + 1}$ for $x \ne 2$ , and $k'(2) = -\frac{1}{3}$ . The gesture for a fraction is the common denominator INSIDE the numerator, then a quotient of fractions turned into a product. The sign of $2 - x$ is where the marks go: dropping it gives $+\frac{1}{3}$ , which contradicts the graph of $k$ , decreasing for $x > -1$ .

e) $m(2) = 8 - 2 = 6$ , so the quotient is $\frac{x^3 - x - 6}{x - 2}$ . Divide: $x^3 - x - 6 = (x - 2)(x^2 + 2x + 3)$ , which checks by expanding, $x^3 + 2x^2 + 3x - 2x^2 - 4x - 6 = x^3 - x - 6$ . The quotient is $x^2 + 2x + 3$ for $x \ne 2$ , and $m'(2) = 4 + 4 + 3 = 11$ . Check: $3x^2 - 1$ at $x = 2$ is $11$ . Every part of this exercise gives the same kind of answer, a finite limit found after a cancellation: that is what differentiable at $a$ means.

Exercise 3: The derivative as a function, by the definition

Letting the point vary turns the derivative into a new function: $f'(x) = \lim_{h\to 0} \frac{f(x + h) - f(x)}{h}$ , defined at every $x$ where this limit exists. The domain of $f'$ can be SMALLER than the domain of $f$ , and saying where the derivative exists is part of the answer.

The algebra is the same as at a point, with $x$ kept as a letter: expand, conjugate, or common denominator, then cancel $h$ , then let $h \to 0$ with $x$ fixed.

a) $f(x) = x^3 - 6x$ : find $f'(x)$ by the definition.
b) $g(x) = \sqrt{4 - x}$ : find $g'(x)$ by the definition, and give the domains of $g$ and of $g'$ .
c) $r(x) = \frac{x}{2x + 1}$ : find $r'(x)$ by the definition, and give the domain of $r'$ .
d) Using a), find the points of the curve $y = x^3 - 6x$ where the tangent is horizontal, then those where it is parallel to the line $y = 6x$ .
e) Explain why $g$ is not differentiable at $x = 4$ by computing the only one-sided quotient that makes sense there, and describe the graph of $g$ at that point.

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Answers

a) $f'(x) = 3x^2 - 6$
b) $g'(x) = -\frac{1}{2\sqrt{4 - x}}$ ; domain of $g$ : $(-\infty, 4]$ , of $g'$ : $(-\infty, 4)$
c) $r'(x) = \frac{1}{(2x + 1)^2}$ , for $x \ne -\frac{1}{2}$
d) Horizontal at $(\sqrt 2, -4\sqrt 2)$ and $(-\sqrt 2, 4\sqrt 2)$ ; parallel to $y = 6x$ at $(2, -4)$ and $(-2, 4)$
e) $\frac{g(4 + h) - g(4)}{h} = -\frac{1}{\sqrt{-h}} \to -\infty$ as $h \to 0^-$ : vertical tangent at the endpoint $(4, 0)$ .

a) $(x + h)^3 = x^3 + 3x^2h + 3xh^2 + h^3$ , so $f(x + h) - f(x) = x^3 + 3x^2h + 3xh^2 + h^3 - 6x - 6h - x^3 + 6x = 3x^2h + 3xh^2 + h^3 - 6h$ . Dividing by $h \ne 0$ : $3x^2 + 3xh + h^2 - 6$ , and letting $h \to 0$ with $x$ fixed, $f'(x) = 3x^2 - 6$ . The $-6h$ is the term students lose: they write $f(x + h) = (x + h)^3 - 6x$ , applying $f$ to half of its argument, and end with $3x^2$ . A check with the power rule, $3x^2 - 6$ , agrees.

b) For $x < 4$ and $h$ small enough that $4 - x - h \ge 0$ : $\frac{\sqrt{4 - x - h} - \sqrt{4 - x}}{h} = \frac{(4 - x - h) - (4 - x)}{h(\sqrt{4 - x - h} + \sqrt{4 - x})} = \frac{-1}{\sqrt{4 - x - h} + \sqrt{4 - x}}$ . As $h \to 0$ this tends to $-\frac{1}{2\sqrt{4 - x}}$ , a finite number as long as $\sqrt{4 - x} \ne 0$ . So $g'(x) = -\frac{1}{2\sqrt{4 - x}}$ on $(-\infty, 4)$ , while $g$ is defined on $(-\infty, 4]$ : the endpoint $4$ belongs to the domain of $g$ but not to that of $g'$ . The minus sign comes from the $-h$ inside the root, and it matches the graph: $g$ decreases.

c) Common denominator: $r(x + h) - r(x) = \frac{x + h}{2x + 2h + 1} - \frac{x}{2x + 1} = \frac{(x + h)(2x + 1) - x(2x + 2h + 1)}{(2x + 2h + 1)(2x + 1)}$ . The numerator is $2x^2 + x + 2xh + h - 2x^2 - 2xh - x = h$ . So $\frac{r(x + h) - r(x)}{h} = \frac{1}{(2x + 2h + 1)(2x + 1)} \to \frac{1}{(2x + 1)^2}$ . The domain of $r'$ is that of $r$ , all $x \ne -\frac{1}{2}$ . When the numerator of a difference quotient collapses to a single $h$ , as here, the computation is right; a numerator with leftover $x^2$ terms means a bracket was dropped.

d) Horizontal tangent: $f'(x) = 0$ , so $3x^2 = 6$ and $x = \pm\sqrt 2$ . Then $f(\sqrt 2) = 2\sqrt 2 - 6\sqrt 2 = -4\sqrt 2$ and $f(-\sqrt 2) = 4\sqrt 2$ : the points $(\sqrt 2, -4\sqrt 2)$ and $(-\sqrt 2, 4\sqrt 2)$ . Parallel to $y = 6x$ : same slope, $3x^2 - 6 = 6$ , so $x^2 = 4$ and $x = \pm 2$ , with $f(2) = 8 - 12 = -4$ and $f(-2) = 4$ : the points $(2, -4)$ and $(-2, 4)$ . The derivative as a FUNCTION is what makes this question possible: one computation of $f'(x)$ answers for every point of the curve at once. The answer is a set of points, so give both coordinates.

e) At $x = 4$ , $g(4) = 0$ and $g(4 + h) = \sqrt{-h}$ exists only for $h \le 0$ : the quotient can only be taken from the left. For $h < 0$ , write $h = -|h|$ : $\frac{\sqrt{-h} - 0}{h} = \frac{\sqrt{|h|}}{-|h|} = -\frac{1}{\sqrt{|h|}}$ , which tends to $-\infty$ as $h \to 0^-$ . The limit is not a finite number, so $g'(4)$ does not exist. On the graph, the half parabola $y = \sqrt{4 - x}$ , lying on its side, arrives at $(4, 0)$ with a VERTICAL tangent. The formula of b) already warned: $-\frac{1}{2\sqrt{4 - x}}$ blows up as $x \to 4^-$ .

Exercise 4: Recognizing a limit as a derivative

Read backwards, the definition says that any limit of the shape $\lim_{h\to 0} \frac{f(a+h) - f(a)}{h}$ or $\lim_{x\to a} \frac{f(x) - f(a)}{x - a}$ IS a derivative. Recognizing that shape is a standard MATH 140 question: identify $f$ and $a$ , and check that the subtracted constant really is $f(a)$ .

Four limits are used below: (i) $\lim_{h\to 0} \frac{(2+h)^5 - 32}{h}$ , (ii) $\lim_{x\to 9} \frac{\sqrt x - 3}{x - 9}$ , (iii) $\lim_{x\to \pi/4} \frac{\tan x - 1}{x - \frac{\pi}{4}}$ and (iv) $\lim_{t\to 1} \frac{t^4 + t - 2}{t - 1}$ .

a) For each of the limits (i) to (iv), give a function $f$ and a number $a$ such that the limit equals $f'(a)$ .
b) Compute limit (i) without any differentiation rule, by expanding $(2+h)^5$ with the binomial coefficients $1, 5, 10, 10, 5, 1$ .
c) Compute limits (ii) and (iv) by algebra alone.
d) Suppose $f$ is differentiable at $3$ with $f'(3) = 4$ . Find $\lim_{h\to 0} \frac{f(3 + 2h) - f(3)}{h}$ and $\lim_{h\to 0} \frac{f(3 - h) - f(3)}{h}$ .
e) Limit (i) is also $F'(0)$ for another function $F$ : give it. Then say which derivative $\lim_{h\to 0} \frac{|h|}{h}$ represents, and whether it exists.

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Answers

a) (i) $x^5$ at $2$ ; (ii) $\sqrt x$ at $9$ ; (iii) $\tan x$ at $\frac{\pi}{4}$ ; (iv) $t^4 + t$ at $1$
b) $80$
c) (ii) $\frac{1}{6}$ ; (iv) $5$
d) $8$ and $-4$
e) $F(x) = (2 + x)^5$ at $0$ ; $\lim \frac{|h|}{h}$ is $f'(0)$ for $f(x) = |x|$ , and it does not exist ( $1$ on the right, $-1$ on the left).

a) (i) $32 = 2^5$ , so with $f(x) = x^5$ and $a = 2$ the quotient is $\frac{f(2 + h) - f(2)}{h}$ . (ii) $3 = \sqrt 9$ : $f(x) = \sqrt x$ , $a = 9$ , in the $x \to a$ form. (iii) $1 = \tan\frac{\pi}{4}$ : $f(x) = \tan x$ , $a = \frac{\pi}{4}$ ; its value needs the derivative of $\tan$ , which is a later chapter, and the question only asks for $f$ and $a$ . (iv) The subtracted constant must be $f(1)$ : with $f(t) = t^4 + t$ , $f(1) = 2$ and the numerator is $f(t) - f(1)$ , so $a = 1$ . The choice $f(t) = t^4 + t - 2$ also works, since then $f(1) = 0$ and the numerator is $f(t) - f(1)$ . What is NOT accepted is $f(t) = t^4 + t$ with no check: the line $f(1) = 2$ is the proof that the shape really is $f(t) - f(a)$ .

b) $(2 + h)^5 = 32 + 5 \cdot 16h + 10 \cdot 8h^2 + 10 \cdot 4h^3 + 5 \cdot 2h^4 + h^5 = 32 + 80h + 80h^2 + 40h^3 + 10h^4 + h^5$ . Subtract $32$ and divide by $h \ne 0$ : $80 + 80h + 40h^2 + 10h^3 + h^4$ , which tends to $80$ . This is the derivative of $x^5$ at $2$ obtained from the definition alone, and the power rule, $5 \cdot 2^4 = 80$ , confirms it. The structure explains the rule: only the term in $h^1$ of the expansion survives the limit, and its coefficient is $5 \cdot 2^4$ .

c) (ii) Conjugate, or better, factor $x - 9 = (\sqrt x - 3)(\sqrt x + 3)$ for $x \ge 0$ : $\frac{\sqrt x - 3}{x - 9} = \frac{1}{\sqrt x + 3}$ for $x \ne 9$ , so the limit is $\frac{1}{3 + 3} = \frac{1}{6}$ . (iv) $t = 1$ is a root of $t^4 + t - 2$ , so $t - 1$ divides it: $t^4 + t - 2 = (t - 1)(t^3 + t^2 + t + 2)$ , as expanding shows. The quotient is $t^3 + t^2 + t + 2$ for $t \ne 1$ , and the limit is $1 + 1 + 1 + 2 = 5$ . Both checks agree with the rules: $\frac{1}{2\sqrt 9} = \frac{1}{6}$ and $4 \cdot 1^3 + 1 = 5$ .

d) Write $k = 2h$ : as $h \to 0$ , $k \to 0$ , and $\frac{f(3 + 2h) - f(3)}{h} = 2 \cdot \frac{f(3 + k) - f(3)}{k} \to 2f'(3) = 8$ . For the second, write $k = -h$ : $\frac{f(3 - h) - f(3)}{h} = -\frac{f(3 + k) - f(3)}{k} \to -f'(3) = -4$ . The denominator must be EXACTLY the increment added to $3$ ; when it is not, rescale. Answering $4$ to both, because the limit looks like a derivative, is the trap of this part and costs it entirely.

e) $F(x) = (2 + x)^5$ has $F(0) = 32$ , so the limit (i) is also $\frac{F(0 + h) - F(0)}{h}$ , that is $F'(0)$ : the pair $(f, a)$ is not unique. And $\frac{|h|}{h} = \frac{|0 + h| - |0|}{h}$ is the difference quotient of $f(x) = |x|$ at $0$ . For $h > 0$ it equals $1$ , for $h < 0$ it equals $-1$ : the one-sided limits are $1$ and $-1$ , different, so the limit does not exist and $|x|$ is not differentiable at $0$ . Recognizing the shape tells you WHICH derivative a limit is; it does not promise that this derivative exists.

Exercise 5: Differentiable or not: corner, cusp, vertical tangent, jump

$f$ is differentiable at $a$ when the two one-sided limits $\lim_{h\to 0^-}$ and $\lim_{h\to 0^+}$ of $\frac{f(a+h) - f(a)}{h}$ exist, are FINITE, and are EQUAL. Four ways to fail: a corner (two different finite slopes), a cusp (slopes $-\infty$ and $+\infty$ ), a vertical tangent (both $+\infty$ , or both $-\infty$ ), and a discontinuity.

The figure shows a function $g$ on $[-5, 5]$ . An open dot is a point that is NOT on the graph, a filled dot a point that is.

a) From the figure, list the numbers in $(-5, 5)$ where $g$ is not continuous and those where $g$ is not differentiable. Give the reason for each.
b) $p(x) = |x^2 - 4|$ : compute the two one-sided derivatives at $x = 2$ and conclude.
c) Study the differentiability of $q(x) = \sqrt[3]{x - 1}$ at $x = 1$ and of $w(x) = \sqrt{|x|}$ at $x = 0$ . Name the defect of each graph.
d) Show that $v(x) = x|x|$ IS differentiable at $0$ , and give $v'(0)$ .
e) Prove that if $f$ is differentiable at $a$ , then $f$ is continuous at $a$ . Use it to justify two of your answers to a) without computing any quotient.

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Answers

a) Not continuous: $-3$ (jump), $3$ (removable). Not differentiable: $-3$ , $-1$ (corner), $1$ (vertical tangent), $3$ .
b) Left: $-4$ ; right: $4$ . Corner: $p$ is not differentiable at $2$ .
c) $q$ : quotient $h^{-2/3} \to +\infty$ on both sides, vertical tangent. $w$ : $+\infty$ on the right, $-\infty$ on the left, cusp.
d) $\frac{v(h) - v(0)}{h} = |h| \to 0$ , so $v'(0) = 0$
e) $f(a+h) - f(a) = \frac{f(a+h) - f(a)}{h} \cdot h \to f'(a) \cdot 0 = 0$ ; a discontinuity at $-3$ and $3$ forbids a derivative there.

a) At $x = -3$ the graph jumps: from the left it arrives at the open dot $(-3, 2)$ , while $g(-3) = 4$ (filled dot), so $\lim_{x\to -3^-} g(x) = 2 \ne g(-3)$ : not continuous, hence not differentiable. At $x = -1$ the graph is continuous but has a CORNER: slope $-1$ on the left segment, a small positive slope on the right, two different one-sided derivatives. At $x = 1$ the curve is continuous and smooth-looking but its tangent is VERTICAL: the slopes of the secants grow without bound. At $x = 3$ there is a hole at $(3, 4)$ and the value $g(3) = 1.5$ sits elsewhere: a removable discontinuity, so no derivative either. Everywhere else in $(-5, 5)$ , $g$ is differentiable. A corner that is hard to see is still a corner: read the two slopes, not the overall look.

b) $p(2) = 0$ . For $x > 2$ , $x^2 - 4 > 0$ and $p(x) = x^2 - 4$ : $\frac{p(2 + h) - p(2)}{h} = \frac{4h + h^2}{h} = 4 + h \to 4$ as $h \to 0^+$ . For $x$ slightly less than $2$ , $x^2 - 4 < 0$ and $p(x) = 4 - x^2$ : the quotient is $\frac{-4h - h^2}{h} = -4 - h \to -4$ as $h \to 0^-$ . Both one-sided limits are finite but DIFFERENT: $p$ has a corner at $(2, 0)$ and $p'(2)$ does not exist. The absolute value is removed by cases BEFORE the limit, with the sign of what is inside checked on each side.

c) $q(1) = 0$ and $\frac{q(1 + h) - q(1)}{h} = \frac{h^{1/3}}{h} = \frac{1}{h^{2/3}} = \frac{1}{(h^{1/3})^2}$ , positive on both sides, so it tends to $+\infty$ as $h \to 0^+$ AND as $h \to 0^-$ : an infinite limit, hence no derivative, and the graph has a vertical tangent at $(1, 0)$ , the line $x = 1$ . For $w$ : $\frac{\sqrt{|h|}}{h}$ equals $\frac{1}{\sqrt h} \to +\infty$ for $h > 0$ and $-\frac{1}{\sqrt{|h|}} \to -\infty$ for $h < 0$ . Opposite infinite slopes: a CUSP at the origin, the graph coming down steeply and going back up steeply. Same verdict, not differentiable, for two different pictures, and the question asks for the picture too.

d) $v(0) = 0$ and $\frac{v(0 + h) - v(0)}{h} = \frac{h|h|}{h} = |h|$ for $h \ne 0$ , which tends to $0$ from both sides. So $v$ is differentiable at $0$ with $v'(0) = 0$ . Compare with $|x|$ : multiplying by $x$ flattens the corner, because the quotient becomes $|h|$ instead of $\frac{|h|}{h}$ . A formula containing an absolute value is not automatically non-differentiable; the quotient decides.

e) For $h \ne 0$ , $f(a + h) - f(a) = \frac{f(a + h) - f(a)}{h} \cdot h$ . As $h \to 0$ , the first factor tends to $f'(a)$ , a finite number by hypothesis, and the second to $0$ ; by the product law, $f(a + h) - f(a) \to f'(a) \cdot 0 = 0$ , that is $\lim_{h\to 0} f(a + h) = f(a)$ : $f$ is continuous at $a$ . The contrapositive is the useful form: NOT continuous implies NOT differentiable. So at $-3$ (jump) and at $3$ (hole) of a), $g$ is not differentiable, with no quotient to compute. The converse is false, and b) is the counterexample: $p$ is continuous at $2$ and not differentiable there.

Exercise 6: Sketching the graph of f' from the graph of f

The value $f'(x)$ is the slope of the tangent to the graph of $f$ at $x$ . To sketch $f'$ , walk along the graph of $f$ and PLOT ITS SLOPES: where $f$ rises, $f'$ is above the axis; where $f$ falls, below; where the tangent is horizontal, $f'$ crosses or touches the axis; where $f$ has a corner, $f'$ is not defined.

The figure shows $f$ on $[-3, 5]$ : a segment from $(-3, 0)$ to $(-1, 4)$ , then the parabola $y = (x - 1)^2$ from $x = -1$ to $x = 3$ , with vertex $(1, 0)$ , then a horizontal segment from $(3, 4)$ to $(5, 4)$ .

a) Give $f'(-2)$ , $f'(0)$ , $f'(1)$ , $f'(2)$ and $f'(4)$ . For the parabola, find its derivative by the definition first.
b) Compute the one-sided derivatives of $f$ at $x = -1$ and at $x = 3$ from the definition, and conclude.
c) Sketch the graph of $f'$ on $(-3, 5)$ , marking what happens at $x = -1$ and $x = 3$ .
d) Give the intervals where $f' > 0$ , where $f' < 0$ , and where $f' = 0$ , and read each one on the graph of $f$ .
e) A classmate draws $f'$ in one stroke, joining the pieces at $x = -1$ and $x = 3$ with vertical segments. Explain the two errors in that sketch.

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a) $f'(-2) = 2$ , $f'(0) = -2$ , $f'(1) = 0$ , $f'(2) = 2$ , $f'(4) = 0$
b) At $-1$ : $2$ on the left, $-4$ on the right. At $3$ : $4$ on the left, $0$ on the right. Two corners: $f'(-1)$ and $f'(3)$ do not exist.
c) $f' = 2$ on $(-3, -1)$ , $f'(x) = 2(x - 1)$ on $(-1, 3)$ , $f' = 0$ on $(3, 5)$ ; open dots at both ends of each piece.
d) $f' > 0$ on $(-3, -1)$ and $(1, 3)$ ; $f' < 0$ on $(-1, 1)$ ; $f' = 0$ at $x = 1$ and on $(3, 5)$ .
e) $f'(-1)$ and $f'(3)$ do not exist, so nothing is plotted there; and a vertical segment would give $f'$ several values at one $x$ .

a) On $(-3, -1)$ , $f$ is a segment rising $4$ for $2$ across: its slope is $2$ , and the tangent to a line is the line itself, so $f'(-2) = 2$ . For the parabola $u(x) = (x - 1)^2 = x^2 - 2x + 1$ : $\frac{u(x + h) - u(x)}{h} = \frac{2xh + h^2 - 2h}{h} = 2x + h - 2 \to 2x - 2$ . So on $(-1, 3)$ , $f'(x) = 2(x - 1)$ : $f'(0) = -2$ , $f'(1) = 0$ (the vertex, horizontal tangent), $f'(2) = 2$ . On $(3, 5)$ , $f$ is constant: $f'(4) = 0$ . Reading a slope on a curve by eye gives an estimate; a formula, even for one piece, gives the exact value, and this question can be answered exactly.

b) At $x = -1$ , $f(-1) = 4$ . For $h < 0$ the point $-1 + h$ is on the segment $y = 2x + 6$ : $\frac{2(-1 + h) + 6 - 4}{h} = \frac{2h}{h} = 2$ . For $h > 0$ it is on the parabola: $\frac{(h - 2)^2 - 4}{h} = \frac{h^2 - 4h}{h} = h - 4 \to -4$ . At $x = 3$ , $f(3) = 4$ : from the left, $\frac{(2 + h)^2 - 4}{h} = \frac{4h + h^2}{h} = 4 + h \to 4$ ; from the right, $\frac{4 - 4}{h} = 0$ . At each point the two one-sided derivatives are finite and DIFFERENT: two corners, and $f$ is not differentiable at $-1$ nor at $3$ , although it is continuous everywhere on $[-3, 5]$ . The formula of each piece must be used on its own side of the joint only.

c) The graph of $f'$ has three pieces, drawn in the figure of the solution: the horizontal segment $y = 2$ on $(-3, -1)$ ; the segment $y = 2(x - 1)$ from $(-1, -4)$ to $(3, 4)$ , a straight line because the derivative of a parabola is linear; and the segment $y = 0$ on $(3, 5)$ . At $x = -1$ the graph of $f'$ has open dots at $(-1, 2)$ and $(-1, -4)$ , at $x = 3$ open dots at $(3, 4)$ and $(3, 0)$ , and no filled dot at all: $f'(-1)$ and $f'(3)$ do not exist. A continuous $f$ can have a derivative with jumps: each corner of $f$ becomes a jump of $f'$ , whose size is the change of slope, $-6$ at $-1$ and $-4$ at $3$ .

d) $f' > 0$ on $(-3, -1)$ and on $(1, 3)$ : there $f$ rises, along the first segment and along the right half of the parabola. $f' < 0$ on $(-1, 1)$ : $f$ falls from $4$ to $0$ . $f' = 0$ at $x = 1$ , the bottom of the parabola, and on the whole interval $(3, 5)$ , where $f$ is flat. The sign of $f'$ is read on the graph of $f$ as up, down or level, never as above or below the axis: $f$ is positive on $(-1, 1)$ while $f'$ is negative there. Confusing the height of $f$ with its slope is the error this exercise is built against.

e) First error: the vertical segments assign values to $f'(-1)$ and $f'(3)$ , which do not exist by b); the sketch must show open dots and leave those two $x$ values empty. Second error: a vertical segment on the graph of $f'$ would give $f'$ infinitely many values at the same $x$ , which no function does, since a derivative is a function and has at most one value at each $x$ . The graph of $f'$ is correctly drawn in three separate strokes. Drawing $f'$ in one stroke because $f$ is drawn in one stroke confuses continuity of $f$ with continuity of $f'$ .

Exercise 7: Estimating a derivative from data, and saying what it means

When a function is known only through measurements, its derivative is estimated with secants. At an inner point of the table, the best estimate is the AVERAGE of the slopes of the two neighbouring secants, one to the left and one to the right. Whatever the method, $f'(a)$ has the units of $f$ divided by the units of the variable, and it is a RATE at an instant, never an amount.

During a spring flood, the level $L(t)$ of a river at a gauging station, in metres, is measured every two hours; $t$ is in hours. The figure plots the six measurements: $L(0) = 2.0$ , $L(2) = 2.4$ , $L(4) = 3.2$ , $L(6) = 3.8$ , $L(8) = 4.0$ , $L(10) = 3.9$ .

a) Compute the average rate of change of $L$ on $[2, 6]$ , with its units, and say which line of the figure it is the slope of.
b) Estimate $L'(4)$ by averaging the slopes of the secants on $[2, 4]$ and $[4, 6]$ . Compare with a).
c) Write one sentence, with units, that a hydrologist could publish about your estimate of $L'(4)$ , and say what it predicts for $L(5)$ .
d) Estimate $L'(8)$ and $L'(10)$ as well as the table allows, and say when the flood peaked.
e) At $t = 8$ , a news site writes: the rate of change of the river is almost zero, so the danger is over. Correct the sentence.

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Answers

a) $\frac{3.8 - 2.4}{4} = 0.35$ m/h, the slope of the dashed secant
b) $\frac{0.4 + 0.3}{2} = 0.35$ m/h: the same number, because the two secants average to the secant on $[2, 6]$
c) At $t = 4$ h the river rises at about $0.35$ m per hour; $L(5) \approx 3.2 + 0.35 = 3.55$ m if the rate stays about the same.
d) $L'(8) \approx \frac{0.1 + (-0.05)}{2} = 0.025$ m/h; $L'(10) \approx -0.05$ m/h (left secant only); peak between $t = 8$ and $t = 10$ .
e) $L'(8) \approx 0$ means the level is at its HIGHEST, about $4.0$ m: the danger is greatest, it is only no longer growing.

a) $\frac{L(6) - L(2)}{6 - 2} = \frac{3.8 - 2.4}{4} = \frac{1.4}{4} = 0.35$ metres per hour. It is the slope of the dashed secant joining $(2, 2.4)$ and $(6, 3.8)$ on the figure. An average rate is a quotient of two differences, with the units of $L$ over the units of $t$ ; giving $1.4$ m, the change of level, answers another question.

b) Left secant, on $[2, 4]$ : $\frac{3.2 - 2.4}{2} = 0.4$ m/h. Right secant, on $[4, 6]$ : $\frac{3.8 - 3.2}{2} = 0.3$ m/h. Average: $L'(4) \approx \frac{0.4 + 0.3}{2} = 0.35$ m/h. It coincides with a), and not by chance: with equal steps, the average of the two slopes is $\frac{1}{2}\left(\frac{L(4) - L(2)}{2} + \frac{L(6) - L(4)}{2}\right) = \frac{L(6) - L(2)}{4}$ , the slope of the symmetric secant on $[2, 6]$ . That symmetric secant is the one whose slope is closest to the tangent's for a smooth curve, which is why it is preferred to a one-sided secant.

c) A sentence with its units: at $t = 4$ hours, the river level is rising at a rate of about $0.35$ metre per hour. Since a derivative is a rate at an instant, it predicts a change only if the rate stays about the same: over the next hour, the level should rise by about $0.35$ m, so $L(5) \approx 3.2 + 0.35 = 3.55$ m. Two words matter in that sentence: rising, which carries the sign, and about, which reminds that $0.35$ is an estimate from secants and not the true $L'(4)$ .

d) At $t = 8$ : left secant $\frac{4.0 - 3.8}{2} = 0.1$ , right secant $\frac{3.9 - 4.0}{2} = -0.05$ , average $L'(8) \approx 0.025$ m/h, almost zero. At $t = 10$ , the last measurement, only the left secant exists: $L'(10) \approx \frac{3.9 - 4.0}{2} = -0.05$ m/h, the level is starting to fall. The sign of the estimated derivative changes between $t = 8$ and $t = 10$ : the flood peaked in that interval, close to $t = 8$ , around $4.0$ m. The table cannot say more precisely when, and an honest answer says so.

e) The news site confuses the RATE with the LEVEL. $L'(8) \approx 0$ says that the level is barely changing at $t = 8$ ; together with d), it says that the river is at or near its highest point, $4.0$ m, the maximum of the whole record. At that moment the danger is at its greatest, not over; what is over is the rise. A correct sentence: the river has stopped rising and is at its peak of about $4$ metres; it should begin to fall in the next hours. The same confusion, value of $f$ against value of $f'$ , turns up in every word problem of the chapter.

Exercise 8: Five statements to correct

Each statement below was written by a student on a MATH 140 assignment, and each is false. Say what is wrong, give a counterexample, and write the correct statement.

a) If $f$ is continuous at $a$ , then $f$ is differentiable at $a$ .
b) A tangent line touches the curve at exactly one point, and it never crosses the curve.
c) For a very small $h$ , say $h = 0.001$ , $f'(a) = \frac{f(a + h) - f(a)}{h}$ .
d) If $f'(a) > 0$ , then $f(a) > 0$ .
e) If the graph of $f$ has a tangent line at the point $(a, f(a))$ , then $f$ is differentiable at $a$ .

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Answers

a) False: $|x - 1|$ at $1$ . True the other way: differentiable implies continuous.
b) False: the tangent $y = 3x - 2$ to $y = x^3$ at $(1, 1)$ meets it again at $(-2, -8)$ ; $y = 0$ crosses $y = x^3$ at $0$ .
c) False: for $x^2$ at $1$ the quotient is $2.001$ , not $2$ . $f'(a)$ is the LIMIT of the quotient.
d) False: $f(x) = x - 5$ has $f'(0) = 1 > 0$ and $f(0) = -5$ . $f'(a) > 0$ says $f$ is rising at $a$ .
e) False: $y = \sqrt[3]{x}$ has the vertical tangent $x = 0$ at the origin, and $f'(0)$ does not exist.

a) FALSE. $f(x) = |x - 1|$ is continuous at $1$ , but its difference quotient $\frac{|h|}{h}$ is $1$ for $h > 0$ and $-1$ for $h < 0$ : the one-sided derivatives differ, there is a corner, and $f'(1)$ does not exist. Correct statement: if $f$ is differentiable at $a$ , then $f$ is continuous at $a$ ; equivalently, a discontinuity forbids a derivative. Continuity is necessary, not sufficient.

b) FALSE on both counts. The tangent to $y = x^3$ at $(1, 1)$ has slope $\lim_{h\to 0} \frac{(1 + h)^3 - 1}{h} = \lim_{h\to 0}(3 + 3h + h^2) = 3$ , so it is $y = 3x - 2$ . It meets the curve where $x^3 - 3x + 2 = 0$ , that is $(x - 1)^2(x + 2) = 0$ : at $x = 1$ , and again at $x = -2$ , the point $(-2, -8)$ . And at the origin the tangent to $y = x^3$ is $y = 0$ (slope $\lim h^2 = 0$ ), which the curve CROSSES, from below to above. Correct statement: the tangent at $P$ is the limit of the secants through $P$ ; it is a LOCAL object, which says how the curve behaves near $P$ and nothing about the rest of the curve. The one-point picture comes from the circle, not from the definition.

c) FALSE. The quotient is the slope of a secant, and it is only an approximation. For $f(x) = x^2$ at $a = 1$ : $\frac{(1.001)^2 - 1}{0.001} = \frac{0.002001}{0.001} = 2.001$ , while $f'(1) = 2$ . The error is exactly $h$ here, since the quotient simplifies to $2 + h$ . Correct statement: $f'(a) = \lim_{h\to 0} \frac{f(a + h) - f(a)}{h}$ , and for a small $h$ the quotient is CLOSE to $f'(a)$ . No single $h$ , however small, gives the derivative; the limit does.

d) FALSE. $f(x) = x - 5$ : its quotient is $\frac{(a + h - 5) - (a - 5)}{h} = 1$ , so $f'(0) = 1 > 0$ , while $f(0) = -5 < 0$ . The value and the slope are independent pieces of information: a road can climb while being below sea level. Correct statement: if $f'(a) > 0$ , the graph of $f$ is RISING at $a$ , whatever the sign of $f(a)$ .

e) FALSE. $f(x) = \sqrt[3]{x}$ : the quotient at $0$ is $\frac{h^{1/3}}{h} = \frac{1}{h^{2/3}} \to +\infty$ , so the secants through the origin become vertical, and the graph HAS a tangent line there, the vertical line $x = 0$ . But a vertical line has no slope, and $f'(0)$ does not exist. Correct statement: $f$ is differentiable at $a$ if and only if the graph has a NON-VERTICAL tangent line at $(a, f(a))$ , whose slope is then $f'(a)$ .

Exercise 9: A stone dropped from a bridge: velocity as a limit of average velocities

A stone is dropped from rest from a bridge whose deck is $44.1$ m above the water. Neglecting air resistance, the distance it has fallen after $t$ seconds is $s(t) = 4.9t^2$ metres, until it reaches the water. Velocity is measured downward.

The average velocity on $[a, a + h]$ is $\frac{s(a + h) - s(a)}{h}$ , and the instantaneous velocity at $a$ is its limit as $h \to 0$ , that is $v(a) = s'(a)$ . In this exercise every velocity comes from that limit: no formula of kinematics is used. The figure shows $s$ and the chord from the origin to $(2, 19.6)$ .

a) When does the stone reach the water? Find its average velocity over the whole fall.
b) Show that the average velocity on $[2, 2 + h]$ is $19.6 + 4.9h$ for $h \ne 0$ , and evaluate it for $h = 1$ , $h = 0.1$ , $h = 0.01$ and $h = -0.1$ .
c) Find the instantaneous velocity at $t = 2$ , then the velocity $v(a)$ at any instant $a$ of the fall, by the definition.
d) Find the velocity with which the stone hits the water, in m/s and in km/h.
e) A student reads the slope of the dashed chord of the figure and answers $9.8$ m/s for the velocity at $t = 2$ . What did the student compute, and at which instant is the stone's velocity actually $9.8$ m/s?

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Answers

a) $t = 3$ s; average velocity $\frac{44.1}{3} = 14.7$ m/s
b) $4.9(4 + h) = 19.6 + 4.9h$ : $24.5$ , $20.09$ , $19.649$ and $19.11$ m/s
c) $v(2) = 19.6$ m/s; $v(a) = 9.8a$ m/s
d) $v(3) = 29.4$ m/s $= 105.84$ km/h, about $106$ km/h
e) The average velocity on $[0, 2]$ , $\frac{19.6}{2} = 9.8$ m/s; the instantaneous velocity is $9.8$ m/s at $t = 1$ s.

a) The stone reaches the water when $s(t) = 44.1$ : $4.9t^2 = 44.1$ , so $t^2 = 9$ and $t = 3$ s, the negative root being rejected since the fall starts at $t = 0$ . Average velocity over the fall: $\frac{s(3) - s(0)}{3 - 0} = \frac{44.1}{3} = 14.7$ m/s. This number describes the whole fall; the stone moves slower than that at the start and faster at the end, and the rest of the exercise measures how much faster.

b) $s(2 + h) - s(2) = 4.9\left[(2 + h)^2 - 4\right] = 4.9(4h + h^2)$ , so for $h \ne 0$ the average velocity is $\frac{4.9(4h + h^2)}{h} = 4.9(4 + h) = 19.6 + 4.9h$ m/s. For $h = 1$ : $24.5$ m/s, the average on $[2, 3]$ . For $h = 0.1$ : $19.6 + 0.49 = 20.09$ m/s. For $h = 0.01$ : $19.6 + 0.049 = 19.649$ m/s. For $h = -0.1$ , the interval $[1.9, 2]$ : $19.6 - 0.49 = 19.11$ m/s. The values from the right decrease toward $19.6$ and the value from the left is below it: the simplified formula shows at a glance where they are heading, which a table alone never proves.

c) $v(2) = \lim_{h\to 0}(19.6 + 4.9h) = 19.6$ m/s. At any instant $a$ in $[0, 3)$ : $\frac{s(a + h) - s(a)}{h} = \frac{4.9(2ah + h^2)}{h} = 4.9(2a + h) \to 9.8a$ . So $v(a) = 9.8a$ m/s: the velocity grows by $9.8$ m/s every second, the acceleration of free fall read off the derivative. The units of $s'(a)$ are those of $s$ over those of $t$ , metres per second; a velocity given without units, or in metres, is marked wrong even when the number is right.

d) The impact happens at $t = 3$ , so the velocity is $v(3) = 9.8 \times 3 = 29.4$ m/s, where the value is the limit of the average velocities on $[3 + h, 3]$ with $h < 0$ , the only side that exists. In km/h: $1$ m/s $= 3.6$ km/h, so $29.4 \times 3.6 = 105.84$ km/h, about $106$ km/h. Compare with a): the impact velocity is exactly twice the average velocity of the fall, $2 \times 14.7 = 29.4$ , a feature of a distance proportional to $t^2$ .

e) The chord joins $(0, 0)$ to $(2, 19.6)$ : its slope $\frac{19.6 - 0}{2 - 0} = 9.8$ m/s is the AVERAGE velocity on $[0, 2]$ , not the velocity at $t = 2$ , which is the slope of the TANGENT, $19.6$ m/s, twice as much. The stone has velocity $9.8$ m/s when $9.8a = 9.8$ , at $a = 1$ s. The error of the student is to take a secant for the tangent: on a curve that bends upward, every secant from an earlier instant is less steep than the tangent at its right end. An average velocity needs two instants; an instantaneous velocity needs a limit.

Exercise 10: A final exam question: cost, marginal cost and what the derivative says in words

A workshop that makes bicycle frames estimates its weekly cost at $C(x) = 1500 + 80\sqrt x$ dollars for $x$ frames, for $0 \le x \le 900$ . The derivative $C'(x)$ is called the MARGINAL COST: the rate at which the cost grows with production, in dollars per frame. It is not a cost, it is a rate.

The figure shows $C$ and the secant between $x = 100$ and $x = 400$ . No calculator: every number is exact, or bracketed by exact numbers.

a) Compute $C(0)$ , $C(100)$ , $C(400)$ , and the average rate of change of $C$ on $[100, 400]$ , with units. What does $C(0)$ represent?
b) Find $C'(x)$ by the definition, then $C'(100)$ and $C'(400)$ , with units.
c) Write the cost of the $101$ st frame, $C(101) - C(100)$ , as $\frac{80}{\sqrt{101} + 10}$ , and use $10.04^2 < 101 < 10.05^2$ to show that it lies between $3.99$ and $4$ dollars. Compare with $C'(100)$ .
d) Interpret $C'(400) = 2$ in one sentence, then explain in words why the marginal cost decreases as production grows.
e) The manager writes: the marginal cost at $100$ frames is $4$ dollars, so $100$ frames cost $400$ dollars; and it equals the average cost per frame. Correct both claims, computing the average cost $\frac{C(100)}{100}$ .

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Answers

a) $C(0) = 1500$ , $C(100) = 2300$ , $C(400) = 3100$ dollars; average rate $\frac{8}{3} \approx 2.67$ dollars per frame; $C(0)$ is the fixed cost.
b) $C'(x) = \frac{40}{\sqrt x}$ ; $C'(100) = 4$ and $C'(400) = 2$ dollars per frame
c) $3.99 < \frac{80}{\sqrt{101} + 10} < 4$ : the $101$ st frame costs just under $C'(100) = 4$ dollars.
d) At $400$ frames, one more frame adds about $2$ dollars; $\frac{40}{\sqrt x}$ decreases because each extra frame is cheaper at larger scale.
e) $C(100) = 2300$ dollars, not $400$ ; the average cost is $23$ dollars per frame, not $4$ .

a) $C(0) = 1500$ dollars: the cost before a single frame is made, rent and equipment, the FIXED cost. $C(100) = 1500 + 80 \times 10 = 2300$ and $C(400) = 1500 + 80 \times 20 = 3100$ dollars. Average rate on $[100, 400]$ : $\frac{3100 - 2300}{400 - 100} = \frac{800}{300} = \frac{8}{3}$ dollars per frame, about $2.67$ : the slope of the secant of the figure. On average, each of the frames from the $101$ st to the $400$ th adds $\frac{8}{3}$ dollars to the cost.

b) For $x > 0$ : $\frac{C(x + h) - C(x)}{h} = \frac{80(\sqrt{x + h} - \sqrt x)}{h}$ , a $\frac{0}{0}$ form with roots, so multiply by the conjugate: $\frac{80\left((x + h) - x\right)}{h(\sqrt{x + h} + \sqrt x)} = \frac{80}{\sqrt{x + h} + \sqrt x} \to \frac{80}{2\sqrt x} = \frac{40}{\sqrt x}$ . The constant $1500$ cancels in the first subtraction, as a constant always does. So $C'(x) = \frac{40}{\sqrt x}$ dollars per frame, $C'(100) = \frac{40}{10} = 4$ and $C'(400) = \frac{40}{20} = 2$ dollars per frame. At $x = 0$ the quotient $\frac{80}{\sqrt h}$ tends to $+\infty$ : $C'(0)$ does not exist, and the graph starts with a vertical tangent.

c) $C(101) - C(100) = 80(\sqrt{101} - 10) = \frac{80(101 - 100)}{\sqrt{101} + 10} = \frac{80}{\sqrt{101} + 10}$ , the conjugate again. Since $10.04^2 = 100.8016 < 101 < 101.0025 = 10.05^2$ , we have $10.04 < \sqrt{101} < 10.05$ , so $\frac{80}{20.05} < \frac{80}{\sqrt{101} + 10} < \frac{80}{20.04}$ . And $\frac{80}{20.05} > 3.99$ because $3.99 \times 20.05 = 79.9995 < 80$ , while $\frac{80}{20.04} < 4$ because $4 \times 20.04 = 80.16 > 80$ . So the $101$ st frame costs between $3.99$ and $4$ dollars, a hair under $C'(100) = 4$ . The marginal cost is the rate at an instant; the cost of one more unit is the change over one unit. They are close because one frame is a small step for this curve, and they are not equal because the curve bends.

d) At a production of $400$ frames a week, the cost is rising at $2$ dollars per frame: producing one more frame adds about $2$ dollars. The marginal cost $\frac{40}{\sqrt x}$ is decreasing, since $\sqrt x$ grows: the graph of $C$ gets flatter, and each extra frame is cheaper to make than the previous one. In words for a manager: at larger scale, the equipment is better used, material is bought in larger lots, and the cost of one more frame falls. A sentence of interpretation always states the instant ( $x = 400$ ), the rate with its units, and the direction.

e) First claim: $C'(100) = 4$ dollars per frame is a RATE, the extra cost of about one more frame when $100$ are made; multiplying it by $100$ gives nothing meaningful. The cost of $100$ frames is $C(100) = 2300$ dollars. Second claim: the average cost per frame at $100$ frames is $\frac{C(100)}{100} = \frac{2300}{100} = 23$ dollars per frame, almost six times the marginal cost. The average spreads the fixed $1500$ dollars over the frames made; the marginal cost ignores it, since a constant has zero rate of change. A MATH 140 final gives the marks of such a question to the sentences, not to the arithmetic.

Corrected exercises: the derivative as a limit, and as a function (MATH 140)

Course recap

Part A: the basics (/50)

Exercise 1: From the slope of a secant to the slope of the tangent

Exercise 2: The derivative at a point, by the definition, in both forms

Exercise 3: The derivative as a function, by the definition

Exercise 4: Recognizing a limit as a derivative

Exercise 5: Differentiable or not: corner, cusp, vertical tangent, jump

Part B: problems and reasoning (/50)

Exercise 6: Sketching the graph of f' from the graph of f

Exercise 7: Estimating a derivative from data, and saying what it means

Exercise 8: Five statements to correct

Exercise 9: A stone dropped from a bridge: velocity as a limit of average velocities

Exercise 10: A final exam question: cost, marginal cost and what the derivative says in words

See also

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