How do you find the local max and min for #f (x) = x^(3) - 6x^(2) + 5#?

Look at a graph:

graph{x^3-6x^2+5 [-36.65, 55.83, -32.08, 14.18]}

To find the local maxima and minima of $f(x) = x^3 - 6x^2 + 5$, follow these steps:

1. Compute the first derivative of $f(x)$ to find critical points.
2. Set the first derivative equal to zero and solve for $x$ to find critical points.
3. Evaluate the second derivative at these critical points to determine concavity.
4. Use the second derivative test to classify the critical points as local maxima, local minima, or points of inflection.

Let's follow these steps:

1. Find the first derivative: $f'(x) = 3x^2 - 12x$.

2. Set $f'(x)$ equal to zero and solve for $x$ to find critical points: $3x^2 - 12x = 0$. Factor out $3x$: $3x(x - 4) = 0$. So, $x = 0$ and $x = 4$ are critical points.

3. Find the second derivative: $f''(x) = 6x - 12$.

4. Evaluate the second derivative at the critical points: $f''(0) = -12$ and $f''(4) = 12$.

5. Apply the second derivative test:

   • If $f''(c) > 0$, then $f(c)$ has a local minimum at $c$.
   • If $f''(c) < 0$, then $f(c)$ has a local maximum at $c$.
   • If $f''(c) = 0$, the test is inconclusive.

For $x = 0$: $f''(0) = -12 < 0$, so there is a local maximum at $x = 0$. For $x = 4$: $f''(4) = 12 > 0$, so there is a local minimum at $x = 4$.

Thus, the local maximum occurs at $x = 0$ and the local minimum occurs at $x = 4$.