Determine the intervals on which the following function is concave up or concave down. Identify any inflection points.
​f(x)equals=e Superscript negative x squared divided by 18e−x2/18

We are tasked with finding the intervals where the function \( f(x) = e^{-x^2 / 18} \) is concave up, concave down, and identifying any inflection points.

### Step 1: Find the first derivative \( f'(x) \)
We start by differentiating \( f(x) \) with respect to \( x \):

\[
f(x) = e^{-x^2 / 18}
\]

Using the chain rule, the first derivative is:

\[
f'(x) = e^{-x^2 / 18} \cdot \frac{d}{dx}\left(-\frac{x^2}{18}\right)
\]

The derivative of \( -\frac{x^2}{18} \) is \( -\frac{2x}{18} = -\frac{x}{9} \). Therefore,

\[
f'(x) = e^{-x^2 / 18} \cdot \left(-\frac{x}{9}\right)
\]
\[
f'(x) = -\frac{x}{9} e^{-x^2 / 18}
\]

### Step 2: Find the second derivative \( f''(x) \)
Now, we differentiate \( f'(x) \) to find the second derivative:

\[
f'(x) = -\frac{x}{9} e^{-x^2 / 18}
\]

Using the product rule for differentiation:

\[
f''(x) = \frac{d}{dx}\left(-\frac{x}{9}\right) \cdot e^{-x^2 / 18} + \left(-\frac{x}{9}\right) \cdot \frac{d}{dx}\left(e^{-x^2 / 18}\right)
\]

We already know from earlier that the derivative of \( e^{-x^2 / 18} \) is:

\[
\frac{d}{dx}\left(e^{-x^2 / 18}\right) = e^{-x^2 / 18} \cdot \left(-\frac{x}{9}\right)
\]

So, substituting back:

\[
f''(x) = -\frac{1}{9} e^{-x^2 / 18} + \left(-\frac{x}{9}\right) \cdot e^{-x^2 / 18} \cdot \left(-\frac{x}{9}\right)
\]
\[
f''(x) = -\frac{1}{9} e^{-x^2 / 18} + \frac{x^2}{81} e^{-x^2 / 18}
\]
\[
f''(x) = e^{-x^2 / 18} \left(-\frac{1}{9} + \frac{x^2}{81}\right)
\]
\[
f''(x) = e^{-x^2 / 18} \left(\frac{x^2 - 9}{81}\right)
\]

### Step 3: Determine concavity
The concavity of the function depends on the sign of \( f''(x) \).

- If \( f''(x) > 0 \), the function is concave up.
- If \( f''(x) < 0 \), the function is concave down.

Since \( e^{-x^2 / 18} \) is always positive, the sign of \( f''(x) \) depends only on the factor \( \frac{x^2 - 9}{81} \). So, we analyze the sign of \( x^2 - 9 \):

- \( x^2 - 9 > 0 \) when \( |x| > 3 \), which means the function is concave up when \( x < -3 \) or \( x > 3 \).
- \( x^2 - 9 < 0 \) when \( |x| < 3 \), which means the function is concave down when \( -3 < x < 3 \).

### Step 4: Identify inflection points
Inflection points occur when \( f''(x) = 0 \), which happens when \( x^2 - 9 = 0 \). Solving for \( x \):

\[
x^2 = 9 \quad \Rightarrow \quad x = \pm 3
\]

Thus, the inflection points are at \( x = -3 \) and \( x = 3 \).

### Conclusion
- The function is **concave up** on \( (-\infty, -3) \cup (3, \infty) \).
- The function is **concave down** on \( (-3, 3) \).
- There are **inflection points** at \( x = -3 \) and \( x = 3 \).

Would you like further details on this, or do you have any questions?

Here are 5 related questions:
1. How do you determine concavity from the second derivative?
2. What is the relationship between inflection points and the second derivative?
3. Can a function have multiple inflection points?
4. How does the sign of the second derivative affect the graph of a function?
5. What is the significance of the function \( e^{-x^2 / 18} \) in real-world applications?

**Tip:** Inflection points are where the concavity of a function changes, and they're found where the second derivative is zero or undefined.

Determine the intervals on which the following function is concave up or concave down. Identify any inflection points. f(x) = e^(-x^2 / 18)

Discover how to find the intervals of concavity and identify inflection points for the function f(x) = e^(-x^2 / 18). This detailed solution includes step-by-step derivatives and the application of the concavity test. Ideal for high school and college-level calculus students.

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