the function f given by f(x)=9x^{2/3}+3x - 6 has a relative minimum at x=(a) -8 (b) -\\sqrt3{2} (c) -1 (d)…

Answer:

A. -8

Explanation:

Step1: Find the derivative

$f'(x)=9\times\frac{2}{3}x^{-\frac{1}{3}}+3 = 6x^{-\frac{1}{3}}+3$.

Step2: Set the derivative equal to 0

$6x^{-\frac{1}{3}}+3 = 0$. $6x^{-\frac{1}{3}}=-3$. $x^{-\frac{1}{3}}=-\frac{1}{2}$.

Step3: Solve for x

Raise both sides to the - 3 power: $x = (-2)^{3}=-8$.

Step4: Check the second - derivative

$f''(x)=6\times(-\frac{1}{3})x^{-\frac{4}{3}}=- 2x^{-\frac{4}{3}}$. When $x = - 8$, $f''(-8)=-2\times(-8)^{-\frac{4}{3}}=-2\times\frac{1}{16}=-\frac{1}{8}<0$ at $x=-8$ the function has a relative minimum.