Question

(7) 11. a) find the critical numbers for

$g(x)=12sqrt3{x}-x$.

b) find the absolute maximum and minimum values taken by $g(x)$ on $-1,27$.

Explanation:

Step1: Rewrite the function

Rewrite $g(x)=12x^{\frac{1}{3}}-x$.

Step2: Find the derivative

Using the power - rule $(x^n)' = nx^{n - 1}$, we have $g'(x)=12\times\frac{1}{3}x^{-\frac{2}{3}}-1 = 4x^{-\frac{2}{3}}-1=\frac{4}{x^{\frac{2}{3}}}-1$.

Step3: Find critical numbers

Set $g'(x) = 0$, then $\frac{4}{x^{\frac{2}{3}}}-1 = 0$.
Add 1 to both sides: $\frac{4}{x^{\frac{2}{3}}}=1$.
Cross - multiply to get $x^{\frac{2}{3}} = 4$.
Cube both sides: $x^{2}=64$, so $x=\pm8$. Also, $g'(x)$ is undefined when $x = 0$. So the critical numbers are $x=-8,0,8$. But for the domain of the original function (since we have a cube - root, the domain is all real numbers), and considering the context of part (b) where we are looking at the interval $[-1,27]$, the relevant critical numbers in the interval $[-1,27]$ are $x = 0,8$.

Step4: Evaluate the function at critical numbers and endpoints

Evaluate $g(x)$ at $x=-1,0,8,27$.
$g(-1)=12(-1)^{\frac{1}{3}}-(-1)=-12 + 1=-11$.
$g(0)=12(0)^{\frac{1}{3}}-0 = 0$.
$g(8)=12(8)^{\frac{1}{3}}-8=12\times2 - 8=16$.
$g(27)=12(27)^{\frac{1}{3}}-27=12\times3 - 27=9$.

Answer:

a) The critical numbers in the interval $[-1,27]$ are $x = 0,8$.
b) The absolute maximum value of $g(x)$ on $[-1,27]$ is $16$ at $x = 8$, and the absolute minimum value is $-11$ at $x=-1$.