Question

let

$f(x)=6(x - 6)^{2/3}+6$

(a) find the interval(s) where $f(x)$ is increasing.

(b) find the interval(s) where $f(x)$ is decreasing.

(c) find the $x$-value(s) of all relative maxima of $f(x)$. if there are none, enter none. if there are multiple relative maxima, separate the values with commas.

$x =$

(d) find the $x$-value(s) of all relative minima of $f(x)$. if there are none, enter none. if there are multiple relative minima, separate the values with commas.

$x =$

Explanation:

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

Use the chain - rule. If \(y = 6(u)^{2/3}+6\) where \(u=x - 6\), then \(\frac{dy}{du}=6\times\frac{2}{3}u^{-1/3}=4u^{-1/3}\) and \(\frac{du}{dx}=1\). So \(f^\prime(x)=\frac{4}{(x - 6)^{1/3}}\).

Step2: Find the critical points

Set \(f^\prime(x) = 0\). Since \(\frac{4}{(x - 6)^{1/3}}=0\) has no solution (the numerator is non - zero), the only place where \(f^\prime(x)\) is undefined is when \(x = 6\).

Step3: Test the intervals

Choose test points in the intervals \((-\infty,6)\) and \((6,\infty)\). Let's choose \(x = 5\) for the interval \((-\infty,6)\) and \(x = 7\) for the interval \((6,\infty)\).
When \(x = 5\), \(f^\prime(5)=\frac{4}{(5 - 6)^{1/3}}=-4<0\).
When \(x = 7\), \(f^\prime(7)=\frac{4}{(7 - 6)^{1/3}}=4>0\).

(a)

Since \(f^\prime(x)>0\) for \(x>6\), the function \(f(x)\) is increasing on the interval \((6,\infty)\).

(b)

Since \(f^\prime(x)<0\) for \(x<6\), the function \(f(x)\) is decreasing on the interval \((-\infty,6)\).

(c)

Since the function changes from decreasing to increasing at \(x = 6\), there is no relative maximum. So \(x=\text{NONE}\).

(d)

Since the function changes from decreasing to increasing at \(x = 6\), the relative minimum occurs at \(x = 6\).

Answer:

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