Corrected exercises: the derivative at a point and as a function (MATH 203)

$f(-1) = 4 + 3 - 1 = 6$. Replace EVERY $x$ by $(-1 + h)$, with brackets: $f(-1 + h) = 4 - 3(-1 + h) - (-1 + h)^2$. The square is $(-1 + h)^2 = 1 - 2h + h^2$, and the minus sign in front of it applies to ALL THREE terms: $-(1 - 2h + h^2) = -1 + 2h - h^2$. So $f(-1 + h) = 4 + 3 - 3h - 1 + 2h - h^2 = 6 - h - h^2$. The constants cancel in $f(-1 + h) - f(-1) = -h - h^2$, as they always must. For $h \ne 0$, $\frac{-h - h^2}{h} = -1 - h$, and $f'(-1) = \lim_{h\to 0}(-1 - h) = -1$. This is the gesture where MATH 203 marks go: the calculus is one line, the minus sign in front of a bracket is the whole difficulty.

$\frac{f(x) - f(-1)}{x - (-1)} = \frac{4 - 3x - x^2 - 6}{x + 1} = \frac{-x^2 - 3x - 2}{x + 1}$. Factor out the minus sign first: $-x^2 - 3x - 2 = -(x^2 + 3x + 2) = -(x + 1)(x + 2)$. For $x \ne -1$ the quotient is $-(x + 2)$, and $f'(-1) = \lim_{x\to -1} -(x + 2) = -1$, as in a). The root test is the safety net of this form: $x = -1$ always makes the numerator $0$, so $(x + 1)$ always divides it. If it does not, $f(-1)$ was computed wrongly.

The tangent passes through $P(-1, 6)$ with slope $-1$: $y - 6 = -1\,(x + 1)$, that is $y = 5 - x$. Check: $x = -1$ gives $y = 6$.

$L_2$ has slope $-1$ (from $(-3, 8)$ to $(2, 3)$ on the grid): it is the tangent $y = 5 - x$, and $4 - 3x - x^2 = 5 - x$ gives $x^2 + 2x + 1 = (x + 1)^2 = 0$, a DOUBLE root, the algebraic signature of a tangent. $L_1$ is $y = 3 - 3x$, slope $-3$: $4 - 3x - x^2 = 3 - 3x$ gives $x^2 = 1$, so it meets the curve again at $x = 1$, the point $(1, 0)$. $L_1$ is the secant $PQ$ with $Q(1, 0)$, and slope $-3$ is exactly what a student gets by writing $(-1 + h)^2 = 1 + h^2$: the middle term $-2h$ lost, the answer is the slope of a secant. $L_3$ is $y = 7 + x$, slope $1$: $4 - 3x - x^2 = 7 + x$ gives $x^2 + 4x + 3 = (x + 1)(x + 3) = 0$, second point $x = -3$, that is $(-3, 4)$. Every line through $P$ other than the tangent cuts the parabola a second time; only the tangent gives a double root. A student who distributes the minus sign to the first term only, $-(-1 + h)^2 = -1 - 2h + h^2$, gets $f(-1 + h) = 6 - 5h + h^2$ and the slope $-5$ of yet another secant.

$f(x) - f(a) = -3(x - a) - (x^2 - a^2) = -3(x - a) - (x - a)(x + a) = -(x - a)(3 + x + a)$. For $x \ne a$ the quotient is $-(3 + x + a)$, so $f'(a) = -(3 + 2a) = -3 - 2a$. Check: $a = -1$ gives $-1$. A horizontal tangent needs $f'(a) = 0$, so $a = -1.5$, and $f(-1.5) = 4 + 4.5 - 2.25 = 6.25$: the vertex $(-1.5, 6.25)$ of the parabola. The difference of squares $x^2 - a^2 = (x - a)(x + a)$ is the one factoring the $x \to a$ form uses all the time.

$g(4) = \sqrt{16} = 4$ and $g(4 + h) = \sqrt{3(4 + h) + 4} = \sqrt{16 + 3h}$. The quotient $\frac{\sqrt{16 + 3h} - 4}{h}$ is a $\frac{0}{0}$ form with a root: multiply by the conjugate $\sqrt{16 + 3h} + 4$. The numerator becomes $(16 + 3h) - 16 = 3h$, so for $h \ne 0$ the quotient is $\frac{3h}{h(\sqrt{16 + 3h} + 4)} = \frac{3}{\sqrt{16 + 3h} + 4}$, and $g'(4) = \frac{3}{4 + 4} = \frac{3}{8}$. Two slips cost marks here: writing $g(4 + h) = \sqrt{3 \cdot 4 + 4} + h$, with $h$ outside the root, and multiplying out the conjugate in the denominator, which destroys the factor that is about to become $8$.

$k(3) = \frac{4}{2} = 2$ and $k(3 + h) = \frac{4 + h}{2 + h}$. Common denominator: $\frac{4 + h}{2 + h} - 2 = \frac{4 + h - 2(2 + h)}{2 + h} = \frac{4 + h - 4 - 2h}{2 + h} = \frac{-h}{2 + h}$. The numerator is a multiple of $h$, which is the sign that the algebra is right. Dividing by $h$ multiplies the denominator by $h$: $\frac{-h}{h(2 + h)} = \frac{-1}{2 + h}$ for $h \ne 0$, so $k'(3) = -\frac{1}{2}$. The negative sign agrees with the graph: $k(x) = 1 + \frac{2}{x - 1}$ decreases for $x > 1$.

$m(4) = \frac{4}{\sqrt{16}} = 1$ and $m(4 + h) = \frac{4}{\sqrt{16 + h}}$. First gesture, common denominator: $\frac{4}{\sqrt{16 + h}} - 1 = \frac{4 - \sqrt{16 + h}}{\sqrt{16 + h}}$. The numerator still has a root and still tends to $0$: second gesture, the conjugate $4 + \sqrt{16 + h}$, which gives $\frac{16 - (16 + h)}{\sqrt{16 + h}\,(4 + \sqrt{16 + h})} = \frac{-h}{\sqrt{16 + h}\,(4 + \sqrt{16 + h})}$. Divide by $h$: the quotient is $\frac{-1}{\sqrt{16 + h}\,(4 + \sqrt{16 + h})}$, which tends to $\frac{-1}{4 \times 8} = -\frac{1}{32}$. Doing the conjugate first, on the whole fraction, is legal but produces a fraction of fractions three storeys high; the order common denominator, then conjugate, keeps every line readable. Check with the calculator: $\frac{m(4.001) - m(4)}{0.001} \approx -0.03125$.

Through $(3, k(3)) = (3, 2)$ with slope $-\frac{1}{2}$: $y - 2 = -\frac{1}{2}(x - 3)$, so $y = -\frac{1}{2}x + \frac{7}{2}$.

In $\frac{4 + h}{2 + h} - 2$, the $2$ must be written over the denominator $2 + h$ before subtracting: $2 = \frac{2(2 + h)}{2 + h} = \frac{4 + 2h}{2 + h}$. The student subtracted the bare $2$ from the numerator, as if $2 = \frac{2}{2 + h}$. Correct numerator: $4 + h - 4 - 2h = -h$, not $2 + h$. The student's line had a warning in it: the numerator $2 + h$ does not tend to $0$, and a difference quotient ALWAYS has a numerator tending to $0$ when $f$ is continuous at $a$. A quotient that blows up like $\frac{1}{h}$ for a function as smooth as $k$ near $3$ is an algebra error, not a discovery.

Common denominator: $\frac{1}{(x + h)^2} - \frac{1}{x^2} = \frac{x^2 - (x + h)^2}{x^2 (x + h)^2}$. The numerator, with the minus sign distributed over the whole square, is $x^2 - x^2 - 2xh - h^2 = -2xh - h^2 = -h(2x + h)$. Dividing by $h \ne 0$: $\frac{-(2x + h)}{x^2 (x + h)^2}$, which tends to $\frac{-2x}{x^4} = -\frac{2}{x^3}$ for $x \ne 0$. So $p'(x) = -\frac{2}{x^3}$ and $p'(2) = -\frac{2}{8} = -\frac{1}{4}$. The typical wrong numerator is $x^2 - x^2 - 2xh + h^2$: the minus sign reached the first two terms only, and the answer is still $-\frac{2}{x^3}$ at the limit by luck, which is why the SIGN must be checked on the line, not on the result.

$\frac{\sqrt{(x + h)^2 + 9} - \sqrt{x^2 + 9}}{h}$: conjugate. The numerator becomes $(x + h)^2 + 9 - (x^2 + 9) = 2xh + h^2 = h(2x + h)$, so for $h \ne 0$ the quotient is $\frac{2x + h}{\sqrt{(x + h)^2 + 9} + \sqrt{x^2 + 9}}$, which tends to $\frac{2x}{2\sqrt{x^2 + 9}} = \frac{x}{\sqrt{x^2 + 9}}$. Since $x^2 + 9 \ge 9 > 0$, this exists for every real $x$: the domain of $q'$ is all of $\mathbb{R}$. And $q'(4) = \frac{4}{\sqrt{25}} = \frac{4}{5}$. The slip to avoid: $\sqrt{(x + h)^2 + 9} = \sqrt{x^2 + 9} + h$, which is false (try $x = 0$, $h = 4$: $5 \ne 7$).

$r(x + h) - r(x) = \frac{(2x + 2h - 1)(x + 3) - (2x - 1)(x + h + 3)}{(x + h + 3)(x + 3)}$. Expand each product in full: $(2x + 2h - 1)(x + 3) = 2x^2 + 5x - 3 + 2hx + 6h$ and $(2x - 1)(x + h + 3) = 2x^2 + 5x - 3 + 2hx - h$. Their difference is $6h - (-h) = 7h$: everything else cancels, as it must. So $\frac{r(x + h) - r(x)}{h} = \frac{7}{(x + h + 3)(x + 3)} \to \frac{7}{(x + 3)^2}$. The domain of $r'$ is that of $r$, all $x \ne -3$, and $r'(-1) = \frac{7}{4}$. Subtracting the second product with the minus sign on its first three terms only, $- 2x^2 - 5x + 3 + 2hx - h$, leaves $4hx + 5h$ in place of $7h$ in the numerator: the missing brackets change the answer.

Slope $16$: $-\frac{2}{x^3} = 16$, so $x^3 = -\frac{1}{8}$ and $x = -\frac{1}{2}$. Then $p\left(-\frac{1}{2}\right) = \frac{1}{1/4} = 4$: the point $\left(-\frac{1}{2}, 4\right)$. Only one point: a cube has one real cube root. A positive slope for $x < 0$ agrees with the graph, which rises toward the asymptote $x = 0$ on the left.

The line $y = 7x$ has slope $7$, so $\frac{7}{(x + 3)^2} = 7$, $(x + 3)^2 = 1$, $x + 3 = \pm 1$, $x = -2$ or $x = -4$. Then $r(-2) = \frac{-5}{1} = -5$ and $r(-4) = \frac{-9}{-1} = 9$: the points $(-2, -5)$ and $(-4, 9)$. Taking only the positive square root, $x + 3 = 1$, loses the second point, and it is a full part of the answer.

(i) $\frac{1}{4} = f(4)$ for $f(x) = \frac{1}{x}$, and the denominator is the increment $h$: $f'(4)$. (ii) $81 = 3^4$: $f(x) = x^4$, $a = 3$, in the $x \to a$ form. (iii) $\frac{1}{2} = \cos\frac{\pi}{3}$: $f(\theta) = \cos\theta$, $a = \frac{\pi}{3}$; its VALUE needs the derivative of cosine, which is section 3.5, and the question asks only for $f$ and $a$. Answering $a = \frac{1}{2}$ confuses the point with the value. (iv) The constant has to be read inside the numerator: with $f(t) = t^3 + 3t$, $f(-2) = -8 - 6 = -14$, so $f(t) - f(-2) = t^3 + 3t + 14$ and $a = -2$. Writing the line $f(-2) = -14$ is the proof that the shape really is $f(t) - f(a)$.

Complex fraction: $\frac{1}{4 + h} - \frac{1}{4} = \frac{4 - (4 + h)}{4(4 + h)} = \frac{-h}{4(4 + h)}$. Dividing by $h \ne 0$ multiplies the denominator by $h$: $\frac{-h}{4h(4 + h)} = \frac{-1}{4(4 + h)}$, which tends to $-\frac{1}{16}$. The bracket in $4 - (4 + h)$ carries the sign of the answer: without it the numerator is $4 - 4 + h = h$ and the result comes out positive, although $\frac{1}{x}$ decreases.

(ii) Difference of squares twice: $x^4 - 81 = (x^2 - 9)(x^2 + 9) = (x - 3)(x + 3)(x^2 + 9)$. For $x \ne 3$ the quotient is $(x + 3)(x^2 + 9)$, which tends to $6 \times 18 = 108$. (iv) $t = -2$ is a root of $t^3 + 3t + 14$, so $t + 2$ divides it: $t^3 + 3t + 14 = (t + 2)(t^2 - 2t + 7)$, as expanding confirms ($t^3 - 2t^2 + 7t + 2t^2 - 4t + 14$). The limit is $4 + 4 + 7 = 15$. The rules of section 3.3 will check both in one line: $4 \cdot 27 = 108$ and $3 \cdot 4 + 3 = 15$.

Put $k = -3h$: as $h \to 0$, $k \to 0$, and $\frac{f(2 - 3h) - f(2)}{h} = -3 \cdot \frac{f(2 + k) - f(2)}{k} \to -3f'(2) = 9$. For the second: $\frac{f(x) - f(2)}{x^2 - 4} = \frac{f(x) - f(2)}{x - 2} \cdot \frac{1}{x + 2}$ for $x \ne 2$, and the product law gives $f'(2) \cdot \frac{1}{4} = -\frac{3}{4}$. In both cases the quotient was REWRITTEN until the true difference quotient appeared, times a factor whose limit is known. Answering $-3$ twice, because both look like derivatives, is the trap.

Split the numerator by adding and subtracting $f(2)$: $\frac{f(2 + h) - f(2 - h)}{h} = \frac{f(2 + h) - f(2)}{h} + \frac{f(2 - h) - f(2)}{-h}$. Each term tends to $f'(2)$ (the second with increment $-h$), so the limit is $2f'(2) = -6$. For $F(x) = |x - 2|$: $F(2 + h) - F(2 - h) = |h| - |-h| = 0$, so the symmetric quotient is $0$ for every $h$, and its limit is $0$. Yet $F$ has a corner at $2$, one-sided slopes $1$ and $-1$, and $F'(2)$ does not exist. The symmetric quotient averages the two one-sided slopes; it can exist when they differ, so it is NOT a definition of the derivative. Only the quotient with $f(a)$ in it is.