Question

find the exact location of all the relative and absolute extrema of the function $f(x)=2x^{2}-4x - 6$ with domain $0,3$. f has? at $(x,y)=($ , ) (smallest $x$-value) f has? at $(x,y)=($ , ) (largest $x$-value)

Explanation:

Step1: Find the derivative

Differentiate $f(x)=2x^{2}-4x - 6$ using the power - rule. The derivative $f^\prime(x)=4x - 4$.

Step2: Find critical points

Set $f^\prime(x) = 0$. So, $4x-4 = 0$, which gives $x = 1$.

Step3: Evaluate the function at critical and end - points

Evaluate $f(x)$ at $x = 0$, $x = 1$, and $x = 3$.
When $x = 0$, $f(0)=2(0)^{2}-4(0)-6=-6$.
When $x = 1$, $f(1)=2(1)^{2}-4(1)-6=2 - 4-6=-8$.
When $x = 3$, $f(3)=2(3)^{2}-4(3)-6=18 - 12-6=0$.

Step4: Determine extrema

Since the second - derivative $f^{\prime\prime}(x)=4>0$, the function is concave up. The relative (and absolute) minimum occurs at $x = 1$ and $y=-8$.
The absolute maximum on the domain $[0,3]$ occurs at $x = 3$ and $y = 0$.

Answer:

$f$ has an absolute and relative minimum at $(x,y)=(1,-8)$ (smallest $x$-value)
$f$ has an absolute maximum at $(x,y)=(3,0)$ (largest $x$-value)