f(x)=2(x - 6)^(2/3)+2. (a) find the interval(s) where f(x) is increasing. (b) find the interval(s) where…

Explanation:

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

Using the chain - rule, if $y = 2(u)^{\frac{2}{3}}+2$ and $u=x - 6$, then $\frac{dy}{du}=\frac{4}{3}u^{-\frac{1}{3}}$ and $\frac{du}{dx}=1$. So, $f^\prime(x)=\frac{4}{3}(x - 6)^{-\frac{1}{3}}=\frac{4}{3\sqrt[3]{x - 6}}$.

Step2: Find where $f^\prime(x)=0$ and where it is undefined

The derivative $f^\prime(x)$ is never equal to zero since the numerator $4\neq0$. It is undefined when $x - 6=0$, i.e., $x = 6$.

Step3: Determine intervals of increase and decrease

We consider the intervals $(-\infty,6)$ and $(6,\infty)$. For $x\in(-\infty,6)$, let's take a test - point, say $x = 5$. Then $f^\prime(5)=\frac{4}{3\sqrt[3]{5 - 6}}=-\frac{4}{3}<0$. So, $f(x)$ is decreasing on the interval $(-\infty,6)$. For $x\in(6,\infty)$, let's take a test - point, say $x = 7$. Then $f^\prime(7)=\frac{4}{3\sqrt[3]{7 - 6}}=\frac{4}{3}>0$. So, $f(x)$ is increasing on the interval $(6,\infty)$.

Step4: Find relative maxima and minima

Since $f(x)$ changes from decreasing to increasing at $x = 6$, $f(x)$ has a relative minimum at $x = 6$. There are no relative maxima.

Answer:

(a) $(6,\infty)$ (b) $(-\infty,6)$ (c) NONE (d) $6$