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 ? (smallest $x$-value)
$\boldsymbol{\blacktriangledown}$ at $(x,y)=(\boldsymbol{\square},\boldsymbol{\square})$
$f$ has ? (largest $x$-value)
$\boldsymbol{\blacktriangledown}$ at $(x,y)=(\boldsymbol{\square},\boldsymbol{\square})$
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Explanation:

Step1: Find critical point

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

Step2: Evaluate f at critical point

Calculate $f(1)$:
$f(1) = 3(1)^2 - 6(1) - 9 = 3 - 6 - 9 = -12$

Step3: Evaluate f at domain endpoints

Calculate $f(0)$:
$f(0) = 3(0)^2 - 6(0) - 9 = -9$
Calculate $f(3)$:
$f(3) = 3(3)^2 - 6(3) - 9 = 27 - 18 - 9 = 0$

Step4: Classify extrema

Compare values: $f(1)=-12$, $f(0)=-9$, $f(3)=0$.
The smallest value is $-12$ at $x=1$; the largest value is $0$ at $x=3$.

Answer:

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