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 a relative minimum and an absolute minimum at (x,y)=( ) (smallest x - value) f has an absolute maximum at (x,y)=( ) (largest x - value)

Explanation:

Step1: Find the derivative of the function

The function is $f(x)=3x^{2}-6x - 9$. Using the power - rule $(x^n)'=nx^{n - 1}$, we have $f'(x)=6x-6$.

Step2: Find the critical points

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

Step3: Evaluate the function at the critical point and endpoints

The domain is $[0,3]$. Evaluate $f(x)$ at $x = 0$, $x=1$, and $x = 3$.

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

Step4: Determine relative and absolute extrema

• Since $f'(x)$ changes sign from negative to positive at $x = 1$, $f(x)$ has a relative (and in this case, absolute) minimum at $x = 1$. The point is $(1,-12)$.
• Comparing the function values at the endpoints and the critical point, the absolute maximum occurs at $x = 3$ and the point is $(3,0)$.

Answer:

The relative (and absolute) minimum occurs at $(1,-12)$ and the absolute maximum occurs at $(3,0)$.