Consider the following. (If an answer does not exist, enter DNE.)
f(x) =
1
2
x4 − 4x2 + 2
(a)
Find the interval(s) of increase. (Enter your answer using interval notation.)

Correct: Your answer is correct.
Find the interval(s) of decrease. (Enter your answer using interval notation.)

Correct: Your answer is correct.
(b)
Find the local minimum value(s). (Enter your answers as a comma-separated list.)

Correct: Your answer is correct.
Find the local maximum value(s). (Enter your answers as a comma-separated list.)

Correct: Your answer is correct.
(c)
Find the inflection points.
smaller x-value    (x, y) =

Incorrect: Your answer is incorrect.

larger x-value    (x, y) =

Find the interval(s) where the function is concave up. (Enter your answer using interval notation.)

Find the interval(s) where the function is concave down. (Enter your answer using interval notation.)

Given the function \( f(x) = \frac{1}{2}x^4 - 4x^2 + 2 \), let's solve part (c), which involves finding the inflection points, and the intervals of concavity. To do this, we'll need to compute the second derivative.

### Step 1: First Derivative
We start by differentiating \( f(x) \) to find the critical points.

\[
f'(x) = \frac{d}{dx}\left( \frac{1}{2}x^4 - 4x^2 + 2 \right)
\]
\[
f'(x) = 2x^3 - 8x
\]

### Step 2: Second Derivative
Now, differentiate \( f'(x) \) to get the second derivative.

\[
f''(x) = \frac{d}{dx}(2x^3 - 8x)
\]
\[
f''(x) = 6x^2 - 8
\]

### Step 3: Find Inflection Points
To find the inflection points, we set \( f''(x) = 0 \) and solve for \( x \).

\[
6x^2 - 8 = 0
\]
\[
6x^2 = 8
\]
\[
x^2 = \frac{8}{6} = \frac{4}{3}
\]
\[
x = \pm \sqrt{\frac{4}{3}} = \pm \frac{2\sqrt{3}}{3}
\]

These are the \( x \)-values of the potential inflection points. To get the corresponding \( y \)-values, plug these \( x \)-values into the original function \( f(x) \).

For \( x = \frac{2\sqrt{3}}{3} \):
\[
f\left(\frac{2\sqrt{3}}{3}\right) = \frac{1}{2}\left(\frac{2\sqrt{3}}{3}\right)^4 - 4\left(\frac{2\sqrt{3}}{3}\right)^2 + 2
\]
(Compute similarly for \( x = -\frac{2\sqrt{3}}{3} \))

### Step 4: Intervals of Concavity
To determine the intervals of concavity, we check the sign of \( f''(x) = 6x^2 - 8 \) in different intervals around the inflection points.

- If \( x^2 > \frac{4}{3} \), then \( f''(x) > 0 \), which means the function is concave up.
- If \( x^2 < \frac{4}{3} \), then \( f''(x) < 0 \), which means the function is concave down.

### Final Answer:
- Inflection points: \( \left(\pm \frac{2\sqrt{3}}{3}, f\left(\pm \frac{2\sqrt{3}}{3}\right)\right) \)
- Concave up: \( (-\infty, -\frac{2\sqrt{3}}{3}) \cup (\frac{2\sqrt{3}}{3}, \infty) \)
- Concave down: \( \left(-\frac{2\sqrt{3}}{3}, \frac{2\sqrt{3}}{3}\right) \)

Would you like more details on any specific step?

Here are 5 follow-up questions:
1. How do you verify the nature of concavity for a function?
2. What conditions must hold for a point to be an inflection point?
3. How do critical points differ from inflection points?
4. Can a function have more than two inflection points?
5. How would concavity affect optimization problems?

**Tip:** Inflection points occur where the concavity of a function changes from concave up to concave down or vice versa.

Consider the following. (If an answer does not exist, enter DNE.) f(x) = 1/2 x^4 − 4x^2 + 2. (a) Find the interval(s) of increase. (b) Find the local minimum value(s). (c) Find the inflection points and intervals of concavity.

Explore how to find the inflection points and intervals of concavity for the function f(x) = 1/2 x^4 − 4x^2 + 2. This solution includes detailed steps for computing the first and second derivatives, identifying the inflection points, and analyzing concavity. Suitable for undergraduate calculus students.

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