Question

consider the function f(x)=5 - 6x^2, - 5≤x≤1. the absolute maximum value is and this occurs at x equal to. the absolute minimum value is and this occurs at x equal to.

Explanation:

Step1: Find the derivative

The derivative of $f(x)=5 - 6x^{2}$ is $f^\prime(x)=-12x$.

Step2: Find critical points

Set $f^\prime(x) = 0$, so $-12x=0$, which gives $x = 0$.

Step3: Check endpoints and critical - point

Evaluate $f(x)$ at the endpoints $x=- 5$ and $x = 1$ and the critical - point $x = 0$.
$f(-5)=5-6\times(-5)^{2}=5 - 150=-145$.
$f(0)=5-6\times0^{2}=5$.
$f(1)=5-6\times1^{2}=5 - 6=-1$.

Step4: Determine maximum and minimum

The absolute maximum value is $5$ which occurs at $x = 0$.
The absolute minimum value is $-145$ which occurs at $x=-5$.

Answer:

The absolute maximum value is $5$ and this occurs at $x = 0$. The absolute minimum value is $-145$ and this occurs at $x=-5$.