To find the local extrema for the function $f(x) = -x^3 + 6x^2$ using the Second Derivative Test, we proceed as follows:

Step 1: Find the First Derivative

Calculate $f'(x)$ to locate the critical points:

$f'(x) = \frac{d}{dx}(-x^3 + 6x^2) = -3x^2 + 12x$

Step 2: Set the First Derivative to Zero

To find critical points, set $f'(x) = 0$:

$-3x^2 + 12x = 0$

Factor out $-3x$:

$-3x(x - 4) = 0$

This gives the critical points $x = 0$ and $x = 4$.

Step 3: Find the Second Derivative

Now, compute the second derivative $f''(x)$ to use in the Second Derivative Test:

$f''(x) = \frac{d}{dx}(-3x^2 + 12x) = -6x + 12$

Step 4: Apply the Second Derivative Test

Evaluate $f''(x)$ at each critical point:

1. At $x = 0$: $f''(0) = -6(0) + 12 = 12$ Since $f''(0) > 0$, there is a local minimum at $x = 0$.

2. At $x = 4$: $f''(4) = -6(4) + 12 = -24 + 12 = -12$ Since $f''(4) < 0$, there is a local maximum at $x = 4$.

Step 5: Find the Function Values at the Critical Points

Finally, substitute these $x$-values back into $f(x)$ to find the $y$-coordinates for the ordered pairs:

• For $x = 0$: $f(0) = -(0)^3 + 6(0)^2 = 0$ So, the local minimum is at $(0, 0)$.

• For $x = 4$: $f(4) = -(4)^3 + 6(4)^2 = -64 + 96 = 32$ So, the local maximum is at $(4, 32)$.

Answer

The local extrema, written as ordered pairs, are:

• Local minimum at $(0, 0)$
• Local maximum at $(4, 32)$

Would you like more details on any of these steps, or have any questions?

1. What is the First Derivative Test, and how does it differ from the Second Derivative Test?
2. Why does the sign of $f''(x)$ at critical points determine the type of extremum?
3. How would the process change if $f(x)$ were a different type of polynomial?
4. What is the significance of the critical points in finding extrema?
5. Can we use the Second Derivative Test if $f''(x) = 0$ at a critical point?

Tip: The Second Derivative Test is useful but inconclusive if $f''(x) = 0$ at a critical point; in such cases, consider the First Derivative Test.