To determine the local maximum and minimum values of $f(x) = \sqrt[3]{2x^2 - 18}$ using the second derivative test, we follow these steps:

1. Find the second derivative of $f(x)$.
2. Use the second derivative test to classify the critical points.

Step 1: Find the Second Derivative

We already have the first derivative: $f'(x) = \frac{4x}{3} (2x^2 - 18)^{-2/3}$

Now, we need to find the second derivative, $f''(x)$.

To differentiate $f'(x)$, we use the quotient rule and the chain rule. Let: $u(x) = 4x$ $v(x) = 3(2x^2 - 18)^{2/3}$

Then: $f'(x) = \frac{u(x)}{v(x)} = \frac{4x}{3(2x^2 - 18)^{2/3}}$

Using the quotient rule $\left( \frac{u}{v} \right)' = \frac{u'v - uv'}{v^2}$:

First, find $u'$ and $v'$: $u'(x) = 4$

For $v(x)$, let $w(x) = (2x^2 - 18)^{2/3}$: $v(x) = 3w(x)$

Then: $w'(x) = \frac{2}{3}(2x^2 - 18)^{-1/3} \cdot 4x = \frac{8x}{3}(2x^2 - 18)^{-1/3}$

So: $v'(x) = 3w'(x) = 3 \cdot \frac{8x}{3}(2x^2 - 18)^{-1/3} = 8x(2x^2 - 18)^{-1/3}$

Now, apply the quotient rule: $f''(x) = \frac{(4)(3(2x^2 - 18)^{2/3}) - (4x)(8x(2x^2 - 18)^{-1/3})}{(3(2x^2 - 18)^{2/3})^2}$

Simplify the numerator: $f''(x) = \frac{12(2x^2 - 18)^{2/3} - 32x^2(2x^2 - 18)^{-1/3}}{9(2x^2 - 18)^{4/3}}$ $f''(x) = \frac{12(2x^2 - 18)^{2/3} - 32x^2(2x^2 - 18)^{-1/3}}{9(2x^2 - 18)^{4/3}}$

Combine the terms in the numerator over a common denominator: $f''(x) = \frac{12(2x^2 - 18) - 32x^2}{9(2x^2 - 18)^{4/3}} (2x^2 - 18)^{-1/3}$ $f''(x) = \frac{12(2x^2 - 18) - 32x^2}{9(2x^2 - 18)^{4/3}}$ $f''(x) = \frac{24x^2 - 216 - 32x^2}{9(2x^2 - 18)^{4/3}}$ $f''(x) = \frac{-8x^2 - 216}{9(2x^2 - 18)^{4/3}}$

Step 2: Use the Second Derivative Test

The critical points we identified earlier are $x = 0$, $x = 3$, and $x = -3$.

1. At $x = 0$: $f''(0) = \frac{-8(0)^2 - 216}{9(2(0)^2 - 18)^{4/3}}$ $f''(0) = \frac{-216}{9(-18)^{4/3}}$

Since $(2x^2 - 18)$ is negative for $x = 0$, $f''(0)$ is not well-defined (indeterminate). Thus, the second derivative test fails at this point. Therefore, $x = 0$ is an inflection point rather than a local maximum or minimum.

2. At $x = 3$: $f''(3) = \frac{-8(3)^2 - 216}{9(2(3)^2 - 18)^{4/3}}$ $f''(3) = \frac{-72 - 216}{9(18)^{4/3}}$ $f''(3) = \frac{-288}{9 \cdot 18^{4/3}}$

Since $18 > 0$, $(18)^{4/3}$ is positive. Thus, $f''(3) < 0$, indicating a local maximum at $x = 3$.

3. At $x = -3$: $f''(-3) = \frac{-8(-3)^2 - 216}{9(2(-3)^2 - 18)^{4/3}}$ $f''(-3) = \frac{-72 - 216}{9(18)^{4/3}}$ $f''(-3) = \frac{-288}{9 \cdot 18^{4/3}}$

Since $f''(-3) < 0$, indicating a local maximum at $x = -3$.

Conclusion

Using the second derivative test:

• There is a local maximum at $x = 3$ and $x = -3$.

Would you like further details or have any questions?

Here are five questions you might ask:

1. How can you simplify the process of finding second derivatives?
2. What is the significance of an inflection point in a graph?
3. How does the second derivative test differ from the first derivative test?
4. What are some common mistakes when applying the second derivative test?
5. How do you interpret the concavity of a function from its second derivative?

Tip: When the second derivative test is inconclusive, consider using the first derivative test to analyze the behavior of the function around the critical points.