5. -/2 points details my notes scalc9 3.3.023. find the local maximum and local minimum values of f using…

Explanation:

Step1: Find the first - derivative

Differentiate $f(x)=2 + 9x^{2}-6x^{3}$ using the power rule. $f'(x)=18x - 18x^{2}=18x(1 - x)$.

Step2: Find the critical points

Set $f'(x)=0$. So $18x(1 - x)=0$. Solving gives $x = 0$ and $x = 1$.

Step3: Find the second - derivative

Differentiate $f'(x)=18x - 18x^{2}$ to get $f''(x)=18-36x$.

Step4: Apply the second - derivative test

For $x = 0$, $f''(0)=18>0$. So $f(x)$ has a local minimum at $x = 0$. $f(0)=2+9(0)^{2}-6(0)^{3}=2$. For $x = 1$, $f''(1)=18 - 36=-18<0$. So $f(x)$ has a local maximum at $x = 1$. $f(1)=2+9(1)^{2}-6(1)^{3}=2 + 9-6=5$.

Answer:

local maximum value: $5$ local minimum value: $2$