find the absolute maximum and minimum values of each function over the indicated interval, and indicate the…

Explanation:

Step1: Find the derivative

Differentiate $f(x)=3x^{4}-2x^{3}$ using the power - rule. The power rule states that if $y = x^{n}$, then $y^\prime=nx^{n - 1}$. $f^\prime(x)=12x^{3}-6x^{2}=6x^{2}(2x - 1)$

Step2: Find the critical points

Set $f^\prime(x)=0$. $6x^{2}(2x - 1)=0$ $6x^{2}=0$ gives $x = 0$ and $2x-1=0$ gives $x=\frac{1}{2}$. These are the critical points.

Step3: Evaluate the function at critical points and endpoints

Evaluate $f(x)$ at $x=-3,x = 0,x=\frac{1}{2},x = 3$.

• When $x=-3$: $f(-3)=3(-3)^{4}-2(-3)^{3}=3\times81+2\times27=243 + 54=297$
• When $x = 0$: $f(0)=3(0)^{4}-2(0)^{3}=0$
• When $x=\frac{1}{2}$: $f(\frac{1}{2})=3(\frac{1}{2})^{4}-2(\frac{1}{2})^{3}=3\times\frac{1}{16}-2\times\frac{1}{8}=\frac{3}{16}-\frac{1}{4}=\frac{3 - 4}{16}=-\frac{1}{16}$
• When $x = 3$: $f(3)=3(3)^{4}-2(3)^{3}=3\times81-2\times27=243-54 = 189$

Answer:

The absolute maximum value is $297$ and it occurs at $x=-3$. The absolute minimum value is $-\frac{1}{16}$ and it occurs at $x=\frac{1}{2}$.