Question

find the exact location of all the relative and absolute extrema of the function $f(x) = 3x^2 - 6x - 9$ with domain $0,3$.
$f$ has ? at $(x,y) = ($ $)$ (smallest $x$-value)
$f$ has ? at $(x,y) = ($ $)$ (largest $x$-value)

Explanation:

Step1: Find critical points

Take derivative: $f'(x) = 6x - 6$
Set $f'(x)=0$: $6x-6=0 \implies x=1$

Step2: Evaluate $f(x)$ at critical point and endpoints

At $x=0$: $f(0)=3(0)^2-6(0)-9=-9$
At $x=1$: $f(1)=3(1)^2-6(1)-9=3-6-9=-12$
At $x=3$: $f(3)=3(3)^2-6(3)-9=27-18-9=0$

Step3: Classify extrema

Compare values: $-12 < -9 < 0$

Answer:

$f$ has an absolute minimum (smallest x-value) at $(x,y)=(0, -9)$
$f$ has an absolute maximum (largest x-value) at $(x,y)=(3, 0)$
$f$ has a relative minimum at $(x,y)=(1, -12)$