Corrected exercises: what derivatives tell us about the shape of a graph (MATH 140)

a) $f$ is a polynomial, so $f'$ exists everywhere and the only critical numbers are the zeros of $f'(x) = 4x^3 + 12x^2 - 16x = 4x(x^2 + 3x - 4) = 4x(x + 4)(x - 1)$: they are $-4$, $0$ and $1$. The sign chart of the solution figure takes each factor on each interval: with test values, $f'(-5) = -120 < 0$, $f'(-1) = 24 > 0$, $f'\left(\frac{1}{2}\right) = -\frac{9}{2} < 0$, $f'(2) = 48 > 0$. So $f$ is decreasing on $(-\infty, -4]$, increasing on $[-4, 0]$, decreasing on $[0, 1]$ and increasing on $[1, \infty)$. At $-4$, $f'$ goes from $-$ to $+$: local minimum $f(-4) = 256 - 256 - 128 + 1 = -127$. At $0$, from $+$ to $-$: local maximum $f(0) = 1$. At $1$, from $-$ to $+$: local minimum $f(1) = 1 + 4 - 8 + 1 = -2$. The question asks for VALUES: the answer is $-127$, $1$ and $-2$, attained at $x = -4$, $0$ and $1$; writing “the maximum is $x = 0$” confuses where and how much, and costs a mark.

b) $g'(x) = 3x^2 - 12x + 12 = 3(x^2 - 4x + 4) = 3(x - 2)^2$. The only critical number is $2$, with $g(2) = 8 - 24 + 24 - 5 = 3$. But $(x - 2)^2 > 0$ for every $x \neq 2$: $g'$ is positive on BOTH sides of $2$, it does not change sign, and by the First Derivative Test $g$ has no local extremum at $2$. Indeed $g(x) = (x - 2)^3 + 3$, the cube curve shifted. As for monotonicity, $g' > 0$ on $(-\infty, 2)$ and on $(2, \infty)$, so $g$ is increasing on $(-\infty, 2]$ and on $[2, \infty)$, two intervals that share the point $2$: hence $g$ is increasing on the whole real line. A derivative that vanishes at an isolated point does not stop a function from increasing.

c) $h$ is defined for every real $x$ (a cube root accepts negative numbers). For $x \neq 0$, $h'(x) = 1 - x^{-2/3} = \frac{x^{2/3} - 1}{x^{2/3}}$, and $h'(0)$ does not exist since $x^{-2/3}$ is undefined there. So the critical numbers are $0$ (derivative undefined) and the zeros of the numerator, $x^{2/3} = 1$, that is $x = 1$ and $x = -1$. The denominator $x^{2/3} = (\sqrt[3]{x})^2$ is positive for $x \neq 0$, so the sign of $h'$ is that of $x^{2/3} - 1$: positive for $|x| > 1$, negative for $0 < |x| < 1$. At $-1$, $+$ to $-$: local maximum $h(-1) = -1 - 3(-1) = 2$. At $1$, $-$ to $+$: local minimum $h(1) = 1 - 3 = -2$. At $0$, $-$ on both sides: no extremum.

d) By the product rule, $k'(x) = 2x e^x + x^2 e^x = x(x + 2)e^x$. The factor $e^x$ is positive for every $x$, so it never changes the sign: the sign of $k'$ is the sign of $x(x + 2)$, positive for $x < -2$, negative on $(-2, 0)$, positive for $x > 0$. Local maximum at $-2$: $k(-2) = 4e^{-2}$; local minimum at $0$: $k(0) = 0$. The answer stays exact, $4e^{-2} = \frac{4}{e^2}$, about $0.54$ if you want to check the figure mentally with $e^2 \approx 7.4$. The minimum $0$ is even the smallest value of $k$ on the whole line, since $x^2 e^x \ge 0$.

e) A critical number is defined as a number of the DOMAIN where $f'$ is zero OR does not exist: $0$ is in the domain of $h$ and $h'(0)$ does not exist, so $0$ is a critical number, and forgetting it is the most common omission of the chapter. It is then classified like the others, by the sign of $h'$: negative just left of $0$ and just right of $0$, so $h$ keeps decreasing through the origin and $0$ is NOT an extremum. Since $h'(x) = 1 - x^{-2/3}$ becomes very negative near $0$, the curve crosses the origin with a vertical tangent. Here the candidate happens to be rejected; for $|x|$ instead of $x^{1/3}$ it would have been a minimum. A candidate is never skipped, it is tested.

a) $f'(x) = 4x^3 - 6x^2 - 24x$ and $f''(x) = 12x^2 - 12x - 24 = 12(x^2 - x - 2) = 12(x - 2)(x + 1)$. The candidates are $-1$ and $2$. The quadratic $12(x - 2)(x + 1)$ is positive outside its roots and negative between them: $f$ is concave upward on $(-\infty, -1)$ and $(2, \infty)$, concave downward on $(-1, 2)$. The sign changes at both candidates and $f$ is a polynomial, hence continuous, so both are inflection points: $f(-1) = 1 + 2 - 12 + 7 = -2$ and $f(2) = 16 - 16 - 48 + 7 = -41$. An inflection POINT is a point of the curve: the answer is $(-1, -2)$ and $(2, -41)$, not just “$x = -1$ and $x = 2$”.

b) $g'(x) = 5x^4 - 20x^3$ and $g''(x) = 20x^3 - 60x^2 = 20x^2(x - 3)$, so $g''(0) = 0$ and $g''(3) = 0$. Now look at the SIGN. The factor $x^2$ is positive on both sides of $0$, so near $0$ the sign of $g''$ is the sign of $20(x - 3)$, negative on both sides: the curve is concave downward before and after $0$, and $(0, 0)$ is NOT an inflection point. At $3$, the factor $x - 3$ changes sign while $x^2 > 0$: concave downward on $(-\infty, 3)$, upward on $(3, \infty)$, and $(3, g(3)) = (3, 243 - 405) = (3, -162)$ is an inflection point. A double root of $f''$, like $x^2$ here or like $x^4$ at the origin, is the classic candidate that fails.

c) $h'(x) = \frac{5}{3}x^{2/3}$, which exists everywhere with $h'(0) = 0$, and $h''(x) = \frac{10}{9}x^{-1/3} = \frac{10}{9x^{1/3}}$ for $x \neq 0$, undefined at $0$. So $0$ is a candidate although $h''(0)$ is not zero: the list of candidates includes the points where $f''$ does not exist. The cube root has the sign of $x$, so $h'' < 0$ for $x < 0$ and $h'' > 0$ for $x > 0$: the concavity changes. Since $h$ is continuous at $0$, the origin IS an inflection point, with a horizontal tangent there. Discarding $0$ because “$h''(0)$ does not exist, so there is nothing to test” loses the only inflection point of the curve.

d) By the chain rule with inner function $-\frac{x^2}{2}$, $\varphi'(x) = -x e^{-x^2/2}$. By the product rule, $\varphi''(x) = -e^{-x^2/2} + x^2 e^{-x^2/2} = (x^2 - 1)e^{-x^2/2}$. The exponential is positive, so $\varphi''$ has the sign of $x^2 - 1$: positive for $|x| > 1$, negative for $|x| < 1$. The concavity changes at $-1$ and at $1$, and $\varphi(\pm 1) = e^{-1/2} = \frac{1}{\sqrt e}$. Inflection points: $\left(-1, e^{-1/2}\right)$ and $\left(1, e^{-1/2}\right)$, at height about $0.61$ on the figure (with $\sqrt e \approx 1.65$), exactly where the curve stops bending down and starts flattening into its tails. These are the points where the bell is steepest.

a) $f'(x) = 6x^2 - 6x - 36 = 6(x^2 - x - 6) = 6(x - 3)(x + 2)$, so the critical numbers are $-2$ and $3$ (a polynomial has no other kind). $f''(x) = 12x - 6$. At $-2$: $f''(-2) = -30 < 0$, so $f$ has a local maximum, $f(-2) = -16 - 12 + 72 + 1 = 45$. At $3$: $f''(3) = 30 > 0$, local minimum, $f(3) = 54 - 27 - 108 + 1 = -80$. The written form the marker expects names the hypothesis of the test at each point: “$f'(3) = 0$ and $f''(3) = 30 > 0$, so by the Second Derivative Test $f$ has a local minimum at $3$”. The mnemonic: $f'' > 0$, concave UP, a cup, a MINIMUM.

b) $s'(x) = \cos x - \sin x = 0$ when $\tan x = 1$, which on $(0, 2\pi)$ gives $x = \frac{\pi}{4}$ and $x = \frac{5\pi}{4}$ (dividing by $\cos x$ is safe: $\cos x = 0$ would force $\sin x = \pm 1 \neq 0$). $s''(x) = -\sin x - \cos x = -s(x)$. At $\frac{\pi}{4}$: $s\left(\frac{\pi}{4}\right) = \frac{\sqrt 2}{2} + \frac{\sqrt 2}{2} = \sqrt 2$, so $s''\left(\frac{\pi}{4}\right) = -\sqrt 2 < 0$: local maximum $\sqrt 2$. At $\frac{5\pi}{4}$: $s = -\sqrt 2$, $s'' = \sqrt 2 > 0$: local minimum $-\sqrt 2$. Here the Second Derivative Test is clearly the faster route: a sign chart of $\cos x - \sin x$ on $(0, 2\pi)$ would need four test values.

c) $p'(x) = 15x^4 - 15x^2 = 15x^2(x^2 - 1) = 15x^2(x - 1)(x + 1)$: critical numbers $-1$, $0$, $1$. $p''(x) = 60x^3 - 30x = 30x(2x^2 - 1)$. At $1$: $p''(1) = 30 > 0$, local minimum $p(1) = 3 - 5 = -2$. At $-1$: $p''(-1) = -30 < 0$, local maximum $p(-1) = 2$. At $0$: $p''(0) = 0$, the test is SILENT. Go back to the sign of $p'$: near $0$, $15x^2 > 0$ and $x^2 - 1 < 0$, so $p' < 0$ on both sides of $0$ and there is no extremum. What happens instead is visible on the solution figure: $p''(x) = 30x(2x^2 - 1)$ is positive just left of $0$ and negative just right, so the origin is an inflection point with a horizontal tangent. The silence of the test covers three different realities: $x^4$ has a minimum at $0$, $-x^4$ a maximum, $x^3$ neither, and all three have $f'(0) = f''(0) = 0$.

d) $q'(x) = \frac{4}{3}x^{1/3}$, so $q'(0) = 0$ and $0$ is a critical number. But $q''(x) = \frac{4}{9}x^{-2/3}$ does not exist at $0$, and the test needs the value $q''(0)$ and a continuous $f''$ near $0$: its hypotheses fail, it cannot be invoked. The First Derivative Test has no such requirement: $x^{1/3}$ has the sign of $x$, so $q'$ goes from negative to positive and $q$ has a local minimum $q(0) = 0$, the lowest point of the curve since $x^{4/3} = (x^{1/3})^4 \ge 0$. The First Derivative Test always applies at a critical number of a continuous function; the second is a shortcut, to be used only when its hypotheses hold and it speaks.

a) Read the SIGN of $f'$: the curve is below the axis on $[-3, -2)$ and above it on $(-2, 1)$ and $(1, 3]$. So $f$ is decreasing on $[-3, -2]$ and increasing on $[-2, 1]$ and on $[1, 3]$, hence on $[-2, 3]$. The critical numbers are the zeros of $f'$, $-2$ and $1$. At $-2$, $f'$ changes from negative to positive: local minimum of $f$. At $1$, $f'(1) = 0$ but $f' > 0$ on both sides: no sign change, no extremum. The point where $f'$ TOUCHES the axis is a candidate that the sign rejects, exactly like $g(x) = (x - 2)^3 + 3$ of Exercise 1.

b) Now read the SLOPE of the curve of $f'$, which is $f''$. The curve of $f'$ rises on $(-3, -1)$, falls on $(-1, 1)$ and rises again on $(1, 3)$. So $f'' > 0$, $f$ concave upward, on $(-3, -1)$ and $(1, 3)$; $f'' < 0$, concave downward, on $(-1, 1)$. The concavity changes at $x = -1$ and at $x = 1$, which are the inflection points of $f$. The rule to remember: the inflection points of $f$ are the local extrema of $f'$, the places where the slope of $f$ stops growing and starts shrinking, or the reverse.

c) The student read the graph of $f'$ as if it were the graph of $f$. At $x = -1$, $f'(-1) = 1 > 0$: $f$ is increasing there, so it cannot have a local maximum (at a local maximum inside the interval, $f'$ would be $0$ or undefined). What the top of the curve of $f'$ says is that the slope of $f$ is the largest around there: $f$ climbs fastest at $x = -1$, its concavity switches from up to down, and $x = -1$ is an inflection point, as found in b). Before reading anything, write on the figure “this is $y = f'(x)$”: it prevents the confusion.

d) $f$ is increasing on $[-2, 3]$ and $-2 < 0 < 1$, so $f(-2) < f(0) < f(1)$: strictly, since $f' > 0$ on $(-2, 1)$. $f$ is decreasing on $[-3, -2]$, so $f(-3) > f(-2)$. Hence $f$ decreases until $x = -2$ and increases afterwards, and $f(x) \ge f(-2)$ for every $x$ of $[-3, 3]$: the smallest value of $f$ on $[-3, 3]$ is reached at $x = -2$. What the figure does NOT allow is to compare $f(-3)$ with $f(3)$: the sign of $f'$ tells the direction of each move, not how far $f$ travels. The solution figure shows one function with exactly this derivative, chosen with $f(-2) = 0$; any vertical shift of it would fit too.

e) At $x = 1$ two things happen at once: $f'(1) = 0$, so the tangent to the graph of $f$ is horizontal, and $f''$ changes sign (the curve of $f'$ goes from falling to rising), so the concavity switches from downward to upward. The graph of $f$ climbs, flattens to a horizontal tangent, and climbs again, bending the other way: an inflection point with a horizontal tangent, the same shape as $y = x^3$ at the origin. It is a critical number and an inflection point, and not an extremum.