Problem 21 - Calculus A New Horizon

Question

Find: (a) the intervals on which \(f\) is increasing, (b) the intervals on which \(f\) is decreasing, (c) the open intervals on which \(f\) is concave up. (d) the open intervals on which \(f\) is concave down, and (e) the \(x\) -coordinates of all inflection points. $$f(x)=e^{-x^{2} / 2}$$

Step-by-Step Solution

Answer

(a) Increasing on \((-\infty, 0)\), (b) Decreasing on \((0, \infty)\), (c) Concave up on \((-\infty, -1)\) and \((1, \infty)\), (d) Concave down on \((-1, 1)\), (e) Inflection points at \(x = -1\) and \(x = 1\).

Step 1: Find the derivative of the function

First, find the first derivative of the function, \( f(x) = e^{-x^2/2} \). Use the chain rule to differentiate: \[ f'(x) = e^{-x^2/2} \cdot \left(-\frac{x}{2}\right) = -\frac{x}{2} e^{-x^2/2} \]

Step 2: Determine critical points

Set the first derivative equal to zero to find critical points:\[ -\frac{x}{2} e^{-x^2/2} = 0 \]Since \( e^{-x^2/2} eq 0 \) for any real \( x \), the equation simplifies to \( x = 0 \). This is the only critical point.

Step 3: Test intervals for increasing/decreasing

Choose test points in the intervals around the critical point \( x = 0 \) to determine where \( f(x) \) is increasing or decreasing:- For \( x < 0 \), choose \( x = -1 \) to test: \( f'(-1) = -\frac{-1}{2} e^{-(-1)^2/2} = \frac{1}{2} e^{-1/2} > 0 \) (function is increasing)- For \( x > 0 \), choose \( x = 1 \): \( f'(1) = -\frac{1}{2} e^{-1/2} < 0 \) (function is decreasing)Thus, \( f(x) \) is increasing on \( (-\infty, 0) \) and decreasing on \( (0, \infty) \).

Step 4: Find the second derivative

Now, find the second derivative to analyze concavity:Use the product rule on the first derivative:\[ f''(x) = \left(-\frac{1}{2}\right) e^{-x^2/2} + \left(-\frac{x}{2}\right)(-xe^{-x^2/2}) \]Simplifying gives:\[ f''(x) = \left(-\frac{1}{2} + \frac{x^2}{2}\right)e^{-x^2/2} \]

Step 5: Determine intervals of concavity

Set the second derivative equal to zero to find potential inflection points:\[ \left(-\frac{1}{2} + \frac{x^2}{2}\right)e^{-x^2/2} = 0 \]Since \( e^{-x^2/2} eq 0 \), solve:\[ -\frac{1}{2} + \frac{x^2}{2} = 0 \] \[ x^2 = 1 \]So, \( x = \pm 1 \). These are potential inflection points. Test intervals to determine concavity:- For \( x < -1 \), choose \( x = -2 \): \( f''(-2) = \left(-\frac{1}{2} + 2 \right)e^{-2} > 0 \) (concave up)- For \( -1 < x < 1 \), choose \( x = 0 \): \( f''(0) = -\frac{1}{2} < 0 \) (concave down)- For \( x > 1 \), choose \( x = 2 \): \( f''(2) = \left(-\frac{1}{2} + 2 \right)e^{-2} > 0 \) (concave up)Thus, \( f(x) \) is concave up on \( (-\infty, -1) \) and \( (1, \infty) \) and concave down on \( (-1, 1) \).

Step 6: Identify inflection points

The inflection points are where the concavity changes, which occurs at \( x = -1 \) and \( x = 1 \).

Key Concepts

Derivative

To determine the behavior of a function like \(f(x) = e^{-x^2/2}\), we start by finding its derivative. The derivative, denoted as \(f'(x)\), gives us the rate at which the function changes at any given point.
Finding the derivative often involves rules such as the chain rule, product rule, or quotient rule. For \(f(x) = e^{-x^2/2}\), we use the chain rule to differentiate. This results in:
\[ f'(x) = -\frac{x}{2} e^{-x^2/2} \]

Understanding the derivative helps us explore other properties of the function, such as identifying critical points or determining where the function is increasing or decreasing.

Critical Points

Critical points are essential in understanding the function's behavior. A critical point occurs where the first derivative, \(f'(x)\), equals zero or is undefined. These points are where potential changes in the function's direction happen.
For our function:
Set \(-\frac{x}{2} e^{-x^2/2} = 0\). Since \(e^{-x^2/2}\) is never zero, the equation simplifies to \(x = 0\). Thus, the only critical point is at \(x = 0\).

At these critical points, you might find local maximums, minimums, or points of inflection. Knowing the critical points is crucial for sketching the graph and understanding the nature of the function.

Increasing and Decreasing Intervals

To find where the function is increasing or decreasing, we'll analyze the sign of its first derivative \(f'(x)\). If \(f'(x) > 0\), the function is increasing. Conversely, if \(f'(x) < 0\), it's decreasing.
For \(f(x) = e^{-x^2/2}\), examine intervals around the critical point \(x = 0\):

• For \(x < 0\), choose \(x = -1\). Here, \(f'(-1) > 0\), indicating the function is increasing on \((-\infty, 0)\).
• For \(x > 0\), choose \(x = 1\). Here, \(f'(1) < 0\), meaning the function is decreasing on \((0, \infty)\).

An understanding of these intervals helps in sketching how the function behaves over its domain.

Concavity

Concavity tells us about the curvature of the graph. A function can be concave up, resembling a cup, or concave down, resembling a cap. To determine this, we use the second derivative, \(f''(x)\).
The second derivative of \(f(x) = e^{-x^2/2}\) is:
\[ f''(x) = \left(-\frac{1}{2} + \frac{x^2}{2}\right)e^{-x^2/2} \]
Analyze the sign of \(f''(x)\):

• For intervals where \(f''(x) > 0\), the function is concave up.
• For intervals where \(f''(x) < 0\), the function is concave down.

This information helps predict how the slope of a function graph changes — either speeding up (concave up) or slowing down (concave down) in its increase or decrease.

Inflection Points

Inflection points occur where the concavity of a function changes from down to up or vice versa. It is essential for understanding the shape and nature of the graph.
To find inflection points, solve \(f''(x) = 0\). For \(f(x) = e^{-x^2/2}\), solve:
\[ \left(-\frac{1}{2} + \frac{x^2}{2}\right)e^{-x^2/2} = 0 \]
This simplifies to \(x^2 = 1\), resulting in potential inflection points at \(x = -1\) and \(x = 1\).
Test nearby intervals to confirm changes in concavity:

• From \(x < -1\) or \(x > 1\), the sign of \(f''(x) > 0\), concave up.
• Between \(-1 < x < 1\), \(f''(x) < 0\), concave down.

Thus, the function has inflection points at \(x = -1\) and \(x = 1\), marking changes in the graph's curvature.