Questions: Use the Second Derivative Test to find the location of all local extrema in the interval (-6,5) for the function given below. f(x) = (2 x^3)/3 - x^2 - 12 x If there is more than one local maxima or local minima, write each value of x separated by a comma. If a local maxima or local minima does not occur on the function, enter ∅ in the appropriate box. Answer should be exact. The local maxima occur at x= ∅ The local minima occur at x= ∅

Solution Steps

To find the local extrema of the function \( f(x) = \frac{2x^3}{3} - x^2 - 12x \) using the Second Derivative Test, follow these steps:

1. Find the first derivative \( f'(x) \).
2. Find the critical points by setting \( f'(x) = 0 \) and solving for \( x \).
3. Find the second derivative \( f''(x) \).
4. Apply the Second Derivative Test: Evaluate \( f''(x) \) at each critical point.
   • If \( f''(x) > 0 \), the function has a local minimum at that point.
   • If \( f''(x) < 0 \), the function has a local maximum at that point.
5. Check the interval \( (-6, 5) \) to ensure the critical points lie within this range.

The function is given by

\[ f(x) = \frac{2x^3}{3} - x^2 - 12x. \]

To find the critical points, we first compute the first derivative:

\[ f'(x) = 2x^2 - 2x - 12. \]

Next, we set the first derivative equal to zero to find the critical points:

\[ 2x^2 - 2x - 12 = 0. \]

Factoring or using the quadratic formula, we find the critical points:

\[ x = -2, \quad x = 3. \]

Now, we compute the second derivative to apply the Second Derivative Test:

\[ f''(x) = 4x - 2. \]

We evaluate the second derivative at each critical point:

1. For \( x = -2 \):

\[ f''(-2) = 4(-2) - 2 = -8 - 2 = -10 < 0. \]

This indicates that there is a local maximum at \( x = -2 \).

2. For \( x = 3 \):

\[ f''(3) = 4(3) - 2 = 12 - 2 = 10 > 0. \]

This indicates that there is a local minimum at \( x = 3 \).

Final Answer