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 the derivative

The derivative of $f(x)=3x^{2}-6x - 9$ is $f'(x)=6x - 6$.

Step2: Set derivative equal to 0

Set $f'(x)=0$, so $6x - 6 = 0$. Solving for $x$ gives $x = 1$.

Step3: Evaluate function at critical - point and endpoints

Evaluate $f(x)$ at $x = 0$, $x = 1$, and $x = 3$.
$f(0)=3(0)^{2}-6(0)-9=-9$.
$f(1)=3(1)^{2}-6(1)-9=3 - 6 - 9=-12$.
$f(3)=3(3)^{2}-6(3)-9=27-18 - 9 = 0$.

Step4: Determine extrema

The absolute minimum occurs at $(1,-12)$ (smallest $y$ - value) and the absolute maximum occurs at $(3,0)$ (largest $y$ - value).

Answer:

The function $f$ has an absolute minimum at $(x,y)=(1,-12)$ (smallest $x$ - value among minima) and an absolute maximum at $(x,y)=(3,0)$ (largest $x$ - value among maxima).