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.