91Ó°ÊÓ

In calculus, determining where a function is increasing or decreasing is essential for understanding the function's behavior. These intervals are derived from the first derivative of the function. For the function given, \( f(x) = e^{-x^2 / 2} \), we start by finding the first derivative:\[ f'(x) = -xe^{-x^2 / 2} \]The sign of this derivative determines whether the function is increasing or decreasing:

• If \( f'(x) > 0 \), the function is increasing.
• If \( f'(x) < 0 \), the function is decreasing.

For \( -xe^{-x^2 / 2} \), notice that the exponential part \( e^{-x^2 / 2} \) is always positive, since anything raised to an exponential form remains positive. Therefore, the sign of \( f'(x) \) depends on \( -x \):

• If \( -x > 0 \), or equivalently \( x < 0 \), then \( f'(x) > 0 \), making the function increasing.
• If \( -x < 0 \), or equivalently \( x > 0 \), then \( f'(x) < 0 \), making the function decreasing.

Thus, \( f(x) \) is increasing on \( (-\infty, 0) \) and decreasing on \( (0, \infty) \). This pattern reveals a critical shift in behavior at \( x = 0 \), the transition point from increasing to decreasing.

Concavity tells us how the graph of a function curves. It is determined by the second derivative. A function is concave up where its second derivative is positive, and concave down where it is negative. For the function \( f(x) = e^{-x^2 / 2} \), its second derivative is:\[ f''(x) = (x^2 - 1)e^{-x^2 / 2} \]By analyzing the expression \( x^2 - 1 \), we determine the change in concavity:

• If \( x^2 - 1 > 0 \), \( f''(x) > 0 \) indicating concave up.
• If \( x^2 - 1 < 0 \), \( f''(x) < 0 \) indicating concave down.

Solving \( x^2 - 1 > 0 \) gives \(|x| > 1\), i.e., \( (-\infty, -1) \cup (1, \infty) \). In these intervals, the function is concave up.
For \( x^2 - 1 < 0 \), which simplifies to \(-1 < x < 1\), the function is concave down.
Understanding concavity helps us visualize the direction in which the curve is bending, whether it holds a bowl-like shape or forms an arch.

Inflection points occur where the concavity of a function changes from up to down or vice versa. Mathematically, these points appear where the second derivative is zero or undefined, provided there's a sign change.
The function \( f(x) = e^{-x^2 / 2} \) has a second derivative:\[ f''(x) = (x^2 - 1)e^{-x^2 / 2} \]To find where \( f''(x) = 0 \), solve:\[ x^2 - 1 = 0 \]This simplifies to \( x^2 = 1 \), giving solutions \( x = 1 \) and \( x = -1 \). These are the potential inflection points.

• Verify a sign change in \( f''(x) \) around these points to confirm they are true inflection points.

At \( x = -1 \), the solution transitions from concave up \((-\infty, -1)\) to concave down \((-1, 1)\). Similarly, at \( x = 1 \), it changes from concave down \((-1, 1)\) to concave up \((1, \infty)\). Therefore, \( x = -1 \) and \( x = 1 \) are indeed inflection points because of this concavity change.