Questions: Consider the function f(x)=e^-(x-1)^2/18. f(x) has two inflection points at x=C and x=D with C ≤ D where C is and D is For each of the following intervals, determine whether f(x) is concave up or concave down. (-∞, C) Select an answer (C, D): concave up (D, ∞): concave down

Solution Steps

To determine the concavity of the function \( f(x) = e^{-\frac{(x-1)^2}{18}} \), we need to find its second derivative and analyze its sign. The inflection points occur where the second derivative changes sign. We will:

1. Compute the first and second derivatives of \( f(x) \).
2. Find the points where the second derivative is zero (potential inflection points).
3. Determine the sign of the second derivative in the intervals \((-\infty, C)\), \((C, D)\), and \((D, \infty)\).

We start with the function

\[ f(x) = e^{-\frac{(x - 1)^2}{18}}. \]

Next, we compute the first derivative \( f'(x) \) and the second derivative \( f''(x) \):

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

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

To find the inflection points, we set the second derivative \( f''(x) \) equal to zero:

\[ f''(x) = 0. \]

Solving this gives us the inflection points:

\[ x = -2 \quad \text{and} \quad x = 4. \]

We analyze the sign of the second derivative in the intervals defined by the inflection points:

1. For the interval \((-\infty, -2)\):

   • Choose a test point, e.g., \( x = -3 \).
   • \( f''(-3) > 0 \) (concave up).

2. For the interval \((-2, 4)\):

   • Choose a test point, e.g., \( x = 0 \).
   • \( f''(0) < 0 \) (concave down).

3. For the interval \((4, \infty)\):

   • Choose a test point, e.g., \( x = 5 \).
   • \( f''(5) > 0 \) (concave up).

Final Answer