1. find the absolute maximum and absolute minimum values of f(x)=4x^3 - 9x^2 - 6, 1,3

Explanation:

Step1: Find the derivative

Differentiate $f(x)=4x^{3}-9x^{2}-6$ using the power - rule. The derivative $f'(x)=12x^{2}-18x = 6x(2x - 3)$.

Step2: Find the critical points

Set $f'(x)=0$. So $6x(2x - 3)=0$. Solving this gives $x = 0$ and $x=\frac{3}{2}$. But $x = 0$ is not in the interval $[1,3]$, so we only consider $x=\frac{3}{2}$.

Step3: Evaluate the function at critical and end - points

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

• When $x = 1$, $f(1)=4\times1^{3}-9\times1^{2}-6=4 - 9 - 6=-11$.
• When $x=\frac{3}{2}$, $f(\frac{3}{2})=4\times(\frac{3}{2})^{3}-9\times(\frac{3}{2})^{2}-6=4\times\frac{27}{8}-9\times\frac{9}{4}-6=\frac{27}{2}-\frac{81}{4}-6=\frac{54 - 81 - 24}{4}=-\frac{51}{4}=-12.75$.
• When $x = 3$, $f(3)=4\times3^{3}-9\times3^{2}-6=4\times27-9\times9 - 6=108 - 81 - 6=21$.

Answer:

The absolute maximum value is $21$ at $x = 3$ and the absolute minimum value is $-12.75$ at $x=\frac{3}{2}$.