this is similar to section 3.4 problem 30: qianbuscalc2 3.4.030.r for the function g(x)=4 - 5(x + 3)^(2/3)…

Explanation:

Step1: Find the derivative of $g(x)$

Using the chain - rule, if $y = 4-5u^{\frac{2}{3}}$ and $u=x + 3$, then $\frac{dy}{du}=-5\times\frac{2}{3}u^{-\frac{1}{3}}$ and $\frac{du}{dx}=1$. So $g^\prime(x)=-\frac{10}{3}(x + 3)^{-\frac{1}{3}}=-\frac{10}{3\sqrt[3]{x + 3}}$.

Step2: Determine the critical numbers

Critical numbers occur where $g^\prime(x)=0$ or $g^\prime(x)$ is undefined. Since the numerator of $g^\prime(x)$ is non - zero ($- 10\neq0$), $g^\prime(x)$ is never $0$. $g^\prime(x)$ is undefined when the denominator is $0$, i.e., $3\sqrt[3]{x + 3}=0$, which gives $x=-3$. So $x_c=-3$.

Step3: Apply the First - Derivative Test

Choose a test point to the left of $x=-3$, say $x=-4$. Then $g^\prime(-4)=-\frac{10}{3\sqrt[3]{-4 + 3}}=\frac{10}{3}>0$. Choose a test point to the right of $x=-3$, say $x=-2$. Then $g^\prime(-2)=-\frac{10}{3\sqrt[3]{-2+3}}=-\frac{10}{3}<0$. Since the function is increasing ($g^\prime(x)>0$) for $x<-3$ and decreasing ($g^\prime(x)<0$) for $x>-3$, the function has a relative maximum at $x=-3$.

Answer:

(a) $x_c=-3$ (b) Near the critical number $x_c=-3$, when $x<x_c$, the first derivative is positive, and when $x>x_c$, the first derivative is negative. Therefore the function has a relative maximum at $x_c=-3$.