consider the following. (if an answer does not exist, enter dne.) f(x)=x^2/3(x - 4) (a) find the interval(s)…

Explanation:

Step1: Find the derivative

First, use the product - rule $(uv)^\prime = u^\prime v+uv^\prime$, where $u = x^{2/3}$ and $v=x - 4$. The derivative of $u=x^{2/3}$ is $u^\prime=\frac{2}{3}x^{-1/3}$, and $v^\prime = 1$. So $f^\prime(x)=\frac{2}{3}x^{-1/3}(x - 4)+x^{2/3}=\frac{2(x - 4)}{3x^{1/3}}+x^{2/3}=\frac{2x-8 + 3x}{3x^{1/3}}=\frac{5x-8}{3x^{1/3}}$.

Step2: Find the critical points

Set $f^\prime(x)=0$, then $\frac{5x - 8}{3x^{1/3}}=0$. The numerator gives $5x-8 = 0$, so $x=\frac{8}{5}=1.6$. Also, $f^\prime(x)$ is undefined at $x = 0$.

Step3: Test intervals for increasing and decreasing

• Consider the intervals $(-\infty,0)$, $(0,1.6)$ and $(1.6,\infty)$.
  • For $x\in(-\infty,0)$, let $x=-1$. Then $f^\prime(-1)=\frac{5\times(-1)-8}{3\times(-1)^{1/3}}=\frac{-13}{-3}=\frac{13}{3}>0$, so $f(x)$ is increasing on $(-\infty,0)$.
  • For $x\in(0,1.6)$, let $x = 1$. Then $f^\prime(1)=\frac{5\times1-8}{3\times1^{1/3}}=\frac{-3}{3}=-1<0$, so $f(x)$ is decreasing on $(0,1.6)$.
  • For $x\in(1.6,\infty)$, let $x = 2$. Then $f^\prime(2)=\frac{5\times2-8}{3\times2^{1/3}}=\frac{2}{3\times2^{1/3}}>0$, so $f(x)$ is increasing on $(1.6,\infty)$.

Step4: Find local minimum and maximum

• Since $f(x)$ changes from increasing to decreasing at $x = 0$, $f(0)=0^{2/3}(0 - 4)=0$ is a local maximum.
• Since $f(x)$ changes from decreasing to increasing at $x = 1.6$, $f(1.6)=(1.6)^{2/3}(1.6 - 4)\approx1.38\times(-2.4)\approx - 3.31$.

Answer:

(a) $(-\infty,0)\cup(1.6,\infty)$ (b) $(0,1.6)$ (c) local minimum value: $-3.31$ local maximum value: $0$