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

$f(3) = 9 - 12 + 5 = 2$. 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.

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 $1$.

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 $h \to 0$.

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.

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.

$f(-1) = 2 + 3 = 5$. 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.

$\frac{f(x) - f(-1)}{x - (-1)} = \frac{2x^2 - 3x - 5}{x + 1}$. 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 $-7$.

$g(5) = \sqrt 9 = 3$ 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.

$k(2) = 1$. $\frac{k(x) - k(2)}{x - 2} = \frac{\frac{3}{x+1} - 1}{x - 2} = \frac{\frac{3 - (x + 1)}{x + 1}}{x - 2} = \frac{2 - x}{(x + 1)(x - 2)}$. 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 $x > -1$.

$m(2) = 8 - 2 = 6$, 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.

$(x + h)^3 = x^3 + 3x^2h + 3xh^2 + h^3$, 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.

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.

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.

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.

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 $x \to 4^-$.

(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 $f(t) - f(a)$.

$(2 + h)^5 = 32 + 5 \cdot 16h + 10 \cdot 8h^2 + 10 \cdot 4h^3 + 5 \cdot 2h^4 + h^5 = 32 + 80h + 80h^2 + 40h^3 + 10h^4 + h^5$. 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 $5 \cdot 2^4$.

(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 $4 \cdot 1^3 + 1 = 5$.

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.

$F(x) = (2 + x)^5$ 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.