First Derivative Test for Local Extrema

Find the local maxima and minima of $f(x)={\displaystyle \frac{1}{2}}{x}^3-{\displaystyle \frac{3}{2}}{x}^2+1$.

Solution

f'(x)=\frac{3}{2}x^{2}-3x=\frac{3}{2}x(x-2). Critical points: $x=0$ and $x=2$.

Because f' changes from $+$ to $-$ at $x=0$, $f$ has a **local maximum** at $x=0$: $f(0)=1$. Because f' changes from $-$ to $+$ at $x=2$, $f$ has a **local minimum** at $x=2$: $f(2)=4-6+1=-1$.

Find the local maxima and minima of $f(x)=3x^{5/3}-15x^{2/3}$.

Solution

\begin{aligned} f'(x) &= 5x^{2/3}-10x^{-1/3} \\ &= 5x^{-1/3}(x-2) \\ &= \frac{5(x-2)}{\sqrt[3]{x}}. \end{aligned} Critical points: $x=0$ (where f' is undefined) and $x=2$ (where f'=0).

f' changes from $+$ to $-$ at $x=0$: **local maximum** at $x=0$. f' changes from $-$ to $+$ at $x=2$: **local minimum** at $x=2$.

Investigate the local extrema of $f(x)=(x+1{)}^2(x-2{)}^3$.

Solution

\begin{aligned} f'(x) &= 2(x+1)(x-2)^3+(x+1)^2\cdot3(x-2)^2 \\ &= (x+1)(x-2)^2[2(x-2)+3(x+1)] \\ &= (x+1)(x-2)^2(5x-1). \end{aligned} Critical points: $x=-1$, $x=2$, $x={\displaystyle \frac{1}{5}}$.

f' changes from $-$ to $+$ at... wait, checking the sign diagram: - On $(-\mathrm{\infty },-1)$: $(-)(-)(-)=(-)$, so f'<0. - On $(-1,1/5)$: $(+)(-)(-)=(+)$, so f'>0. - On $(1/5,2)$: $(+)(-)(+)=(-)$, so f'<0. - On $(2,\mathrm{\infty })$: $(+)(+)(+)=(+)$, so f'>0. At $x=-1$: sign changes from $-$ to $+$... wait, sign changes from $-$ to $+$, so **local minimum** at $x=-1$. Actually re-checking: at $x=-1$, f' goes from $-$ (on $(-\mathrm{\infty },-1)$) to $+$ (on $(-1,1/5)$), giving a **local minimum** at $x=-1$. At $x=1/5$: f' goes from $+$ to $-$, giving a **local maximum** at $x=1/5$. At $x=2$: f' does not change sign (negative on both sides, except at the point itself where $(x-2{)}^2=0$). Therefore, $f$ has **no local extremum** at $x=2$. In fact, $(2,f(2))$ is a point of inflection.

Investigate the local extrema of $f(x)={\sin }^{3}x+{\cos }^{3}x$.

Solution

Since $f$ is periodic with period $2\pi$, we restrict to $[0,2\pi ]$. The function is differentiable everywhere: f'(x)=3\cos x\sin^2x-3\sin x\cos^2x=3\cos x\sin x(\sin x-\cos x). f'(x)=0 when $\cos x=0$, $\sin x=0$, or $\sin x=\cos x$. Solving in $[0,2\pi ]$: $\cos x=0\Rightarrow x={\displaystyle \frac{\pi }{2}},{\displaystyle \frac{3\pi }{2}}$ $\sin x=0\Rightarrow x=0,\pi ,2\pi$ $\sin x=\cos x\Rightarrow \tan x=1\Rightarrow x={\displaystyle \frac{\pi }{4}},{\displaystyle \frac{5\pi }{4}}$

Critical points: $0,{\displaystyle \frac{\pi }{4}},{\displaystyle \frac{\pi }{2}},\pi ,{\displaystyle \frac{5\pi }{4}},{\displaystyle \frac{3\pi }{2}},2\pi$. To determine the sign of f'(x)=3\cos x\sin x(\sin x-\cos x), we need the signs of $\cos x$, $\sin x$, and $\sin x-\cos x$ in each subinterval. Using test values for $g(x)=\sin x-\cos x$: | Interval | Sign of $\sin x-\cos x$ | |----------|------------------------| | $[0,\pi /4)$ | $-$ (e.g., at $x=\pi /6$: ) | | $(\pi /4,5\pi /4)$ | $+$ (e.g., at $x=\pi /2$: ) | | $(5\pi /4,2\pi ]$ | $-$ (e.g., at $x=3\pi /2$: ) |

From the sign diagram: - Local minima at $x={\displaystyle \frac{\pi }{4}}$, $x=\pi$, and $x={\displaystyle \frac{3\pi }{2}}$. - Local maxima at $x={\displaystyle \frac{\pi }{2}}$, $x={\displaystyle \frac{5\pi }{4}}$, and (by periodicity) at $x=0$ and $x=2\pi$.

The graph of f'(x) is given. Identify the $x$-values of the relative extrema.

Solution

f' is zero at $x=-2$, $x=-1$, and $x=1$, so these are the only critical points. - At $x=-2$: f' changes from $+$ (increasing $f$) to $-$ (decreasing $f$). So $f$ has a **local maximum** at $x=-2$. - At $x=-1$: f' does not change sign (negative on both sides). So $f$ has **no local extremum** at $x=-1$. - At $x=1$: f' changes from $-$ to $+$. So $f$ has a **local minimum** at $x=1$.

Note: this is one of infinitely many functions whose derivative is the given f', because shifting $f$ vertically does not change its derivative.