How do you use the first derivative test to find the local max and local min values of #f(x)= -x^3 + 12x#?

This can be verified by checking the graph of the original function: graph{-x^3+12x [-41.1, 41.07, -20.53, 20.55]}

To use the first derivative test to find the local max and local min values of $f(x) = -x^3 + 12x$:

1. Find the first derivative of the function, $f'(x)$. $f'(x) = -3x^2 + 12$.

2. Set $f'(x)$ equal to zero and solve for $x$ to find critical points. $-3x^2 + 12 = 0$. $x^2 = 4$. $x = \pm 2$.

3. Test the intervals determined by the critical points using the first derivative test:

   • Choose a test point to the left of $x = -2$, say $x = -3$. Plug $x = -3$ into $f'(x)$. $f'(-3) = -3(-3)^2 + 12 = 21$. Since $f'(-3) > 0$, $f(x)$ is increasing to the left of $x = -2$. So, there is a local minimum at $x = -2$.

   • Choose a test point between $x = -2$ and $x = 2$, say $x = 0$. Plug $x = 0$ into $f'(x)$. $f'(0) = -3(0)^2 + 12 = 12$. Since $f'(0) > 0$, $f(x)$ is increasing between $x = -2$ and $x = 2$. So, there is a local maximum at $x = 2$.

   • Choose a test point to the right of $x = 2$, say $x = 3$. Plug $x = 3$ into $f'(x)$. $f'(3) = -3(3)^2 + 12 = -21$. Since $f'(3) < 0$, $f(x)$ is decreasing to the right of $x = 2$. So, there is a local maximum at $x = 2$.

Hence, the local minimum value occurs at $x = -2$ and the local maximum value occurs at $x = 2$.