6.9 First Derivative Test for Local Extrema

Because $f$ is a polynomial, it is differentiable everywhere, so the critical points are the points where $f^{\prime }(x)=0$.
$$f(x)=\frac{1}{2}{x}^3-\frac{3}{2}{x}^2+1$$
$$\Rightarrow f^{\prime }(x)=\frac{3}{2}{x}^2-3x=\frac{3}{2}x(x-2).$$ Critical points: $$f^{\prime }(x)=0⟺x=0,x=2.$$

As this sign table shows, because the sign of $f^{\prime }$ changes from $+$ to $-$ at $x=0$, the function has a local maximum there, and because the sign of $f^{\prime }$ changes from $-$ to $+$ at $x=2$, the function has a local minimum at $x=2$. The graph of $f$ is shown in Figure 4.

First we calculate $f^{\prime }$:
$$\begin{array}{rl}{f}^{\prime }(x)& =3(\frac{5}{3})x^{\frac{5}{3}-1}-15(\frac{2}{3})x^{\frac{2}{3}-1}\\ & =5x^{2/3}-10x^{-1/3}\\ & =5x^{-1/3}(x-2)\\ & =\frac{5(x-2)}{\sqrt[3]{x}}\end{array}$$
It is clear that $f^{\prime }$ is not defined for $x=0$ (because the denominator is zero when $x=0$) and $f^{\prime }(2)=0$. Therefore, the critical points of $f$ are $x=0$ and $x=2$.
Now we need to determine the sign of $f^{\prime }$:

Therefore, $f$ has a local maximum at $x=0$ and a local minimum at $x=2$ and these are the only local extrema. The graph of $f$ is illustrated in Figure 5.

$f$ is a differentiable function everywhere (because it is the product of two differentiable functions $(x+1{)}^2$ and $(x-2{)}^3$). Therefore, the only critical points of $f$ are the zeros of $f^{\prime }$.
$$f(x)=\underset{u}{\underbrace{(x+1{)}^2}}\underset{v}{\underbrace{(x-2{)}^3}}$$
$$\begin{array}{rl}\Rightarrow f^{\prime }(x)& =\underset{u^{\prime }}{\underbrace{2(x+1)}}\underset{v}{\underbrace{(x-2{)}^3}}+\underset{u}{\underbrace{(x+1{)}^2}}\underset{v^{\prime }}{\underbrace{3(x-2{)}^2}}\\ \text{(factor)}& & =(x+1)(x-2{)}^2[2(x-2)+3(x+1)]& =(x+1)(x-2{)}^2(5x-1)\end{array}$$
Critical points: $$f^{\prime }(x)=0⟺x=-1,x=2,x=\frac{1}{5}.$$ Now we need to determine the sign of $f^{\prime }(x)$:

This sign table shows that $f$ has a local maximum at $x=-1$ and a local minimum at $x=-5/4$. Because the sign of $f^{\prime }$ does not change in moving from left to right through $x=1$, $f$ does not have a local extremum there. In fact, $(2,f(2))$ is a point of inflection, as we can see from the graph of $f$ (Figure 6).

This function is periodic with a period $2\pi$ because
$$\begin{array}{rl}f(x+2\pi )& =(\sin (x+2\pi ){)}^3+(\cos (x+2\pi ){)}^3\\ & =(\sin x{)}^3+(\cos x{)}^3\\ & =f(x).\end{array}$$
Therefore, it is sufficient to investigate the local extrema of $f$ in one period, say $[0,2\pi ]$. Note that $f$ is differentiable everywhere and
$$\begin{array}{rl}{f}^{\prime }(x)& =3\cos x{\sin }^{2}x-3\sin x{\cos }^{2}x\\ & =3\cos x\sin x(\sin x-\cos x).\end{array}$$
$f^{\prime }$ is zero wherever $$\cos x=0\text{or}\sin x=0\text{or}\sin x-\cos x=0$$ We find the solutions of the above equations in the interval $[0,2\pi ]$:
$$\cos x=0⟺x=\frac{\pi }{2},x=\frac{3\pi }{2}$$
$$\sin x=0⟺x=0,x=\pi ,x=2\pi$$
$$\sin x-\cos x=0⟺\sin x=\cos x⟺\tan x=\frac{\sin x}{\cos x}=1$$
Using the unit circle (see Figure 7)
$$\tan x=1⟺x=\frac{\pi }{4},x=\pi +\frac{\pi }{4}=\frac{5\pi }{4}$$

Therefore, the critical points of $f$ are $$0,\frac{\pi }{4},\frac{\pi }{2},\pi ,\frac{5\pi }{4},\frac{3\pi }{2},2\pi .$$ Sign of $f^{\prime }(x)$ depends on the signs of $\cos x$, $\sin x$, and $\sin x-\cos x$. It is easy to determine the signs of $\cos x$ and $\sin x$:

The function $g(x)=\sin x-\cos x$ is continuous and might change its sign only at its zeros ($\pi /4$ and $5\pi /4$) [Recall that if $g$ is a continuous function, $g(a)>0$ and $g(b)<0$, then it follows from Bolzano’s Theorem  that $g$ has a zero between $a$ and $b$]. To determine its sign, we choose one test value in each of the intervals $[0,\pi /4]$, $[\pi /4,5\pi /4]$, and $[5\pi /4,2\pi ]$.

Now we can determine the sign of $f^{\prime }(x)$:

From this table, it is clear that $f$ has local minima at $x=\pi /4$, $x=\pi$, and $x=3\pi /2$, and local maxima at $x=\pi /2$ and $x=5\pi /4$. What can we tell about the behavior of the function at $x=0$ and $x=2\pi$? Because $f$ is periodic, the fact that $f$ is decreasing on $[0,\pi /4]$ shows $f$ is also decreasing on $[2\pi ,2\pi +\pi /4$]. Hence, because the behavior of $f$ changes from increasing $<img\; draggable="false"\; role="img"\; class="emoji"\; alt="\nearrow "\; src="https://s.w.org/images/core/emoji/15.0.3/svg/2197.svg">$ to decreasing $<img\; draggable="false"\; role="img"\; class="emoji"\; alt="\searrow "\; src="https://s.w.org/images/core/emoji/15.0.3/svg/2198.svg">$ at $x=2\pi$, $f$ has a local maximum at $x=2\pi$. Analogously we can show $f(0)$ is a local maximum. The graph of $f$ is shown in Figure 8.

$f^{\prime }$ is defined everywhere and is zero at $x=-2,x=-1$, and $x=1$, so these are the only critical points of $f$.

• In moving from left to right through $x=-2$, $f^{\prime }$ changes from positive (increasing $f$ $<img\; draggable="false"\; role="img"\; class="emoji"\; alt="\nearrow "\; src="https://s.w.org/images/core/emoji/15.0.3/svg/2197.svg">$) to negative (decreasing $f$ $<img\; draggable="false"\; role="img"\; class="emoji"\; alt="\searrow "\; src="https://s.w.org/images/core/emoji/15.0.3/svg/2198.svg">$) so $f$ has a local maximum at $x=-2$.
• Because the sign of $f^{\prime }$ does not change at $x=-1$, $f$ does not have a local extremum at $x=-1$.
• Because the sign of $f^{\prime }$ changes from negative (decreasing $f$ $<img\; draggable="false"\; role="img"\; class="emoji"\; alt="\searrow "\; src="https://s.w.org/images/core/emoji/15.0.3/svg/2198.svg">$) to positive (increasing $f$ $<img\; draggable="false"\; role="img"\; class="emoji"\; alt="\nearrow "\; src="https://s.w.org/images/core/emoji/15.0.3/svg/2197.svg">$), $f$ has a minimum at $x=1$.

The graph of $f$ is shown below (we will learn more about how to obtain the graph of $f$ from the graph of $f^{\prime }$ later). Note that this is one graph of infinitely many functions whose derivatives are $f^{\prime }$ because if we shift this graph vertically up or down $y=f(x)+c$, the graph of $f^{\prime }$ remains the same (the derivative of a constant is zero).