consider the function ( f(x)=1 - 6x^{2},-4leq...
consider the function ( f(x)=1 - 6x^{2},-4leq xleq2). the absolute maximum value is and this occurs at ( x = ). the absolute minimum value is and this occurs at ( x = ).
Answer
# Explanation:
## Step1: Find the derivative
Differentiate $f(x)=1 - 6x^{2}$ using the power - rule. The derivative $f^\prime(x)=\frac{d}{dx}(1)-6\frac{d}{dx}(x^{2})=0 - 12x=-12x$.
## Step2: Find critical points
Set $f^\prime(x) = 0$. So, $-12x = 0$, which gives $x = 0$.
## Step3: Evaluate the function at critical and endpoints
Evaluate $f(x)$ at $x=-4,x = 0,x = 2$.
When $x=-4$, $f(-4)=1-6\times(-4)^{2}=1 - 96=-95$.
When $x = 0$, $f(0)=1-6\times0^{2}=1$.
When $x = 2$, $f(2)=1-6\times2^{2}=1 - 24=-23$.
# Answer:
The absolute maximum value is $1$ and this occurs at $x = 0$.
The absolute minimum value is $-95$ and this occurs at $x=-4$.