Problem 37 - Calculus Early Transcendentals

Question

Find the intervals on which \(f\) is increasing and decreasing. $$f(x)=x e^{-x^{2} / 2}$$

Step-by-Step Solution

Answer

Answer: The function is increasing on the interval \((-1, 1)\) and decreasing on the intervals \((-\infty, -1)\) and \((1, \infty)\).

Step 1: Find the first derivative of the function

First, we need to find the derivative of the function \(f(x) = xe^{-x^2/2}\) using the product rule. The product rule states that \((uv)' = u'v + uv'\), where \(u\) and \(v\) are differentiable functions of \(x\). Set \(u = x\) and \(v = e^{-x^2/2}\), then $$u' = 1$$ $$v' = \frac{d}{dx} e^{-x^2/2} = -xe^{-x^2/2}$$ (by applying the chain rule) Now, we can use the product rule to find the derivative \(f'(x)\): $$f'(x) = u'v + uv' = 1 \cdot e^{-x^2/2} + x \cdot (-x)e^{-x^2/2}$$ Simplify the expression: $$f'(x) = e^{-x^2/2} - x^2e^{-x^2/2}$$

Step 2: Find the critical points

Now we need to find the critical points of the function, which occur when \(f'(x) = 0\) or \(f'(x)\) is not defined. In this case, \(f'(x)\) is defined for all x, so we only need to find where \(f'(x)=0\). Set \(f'(x)=0\): $$e^{-x^2/2} - x^2e^{-x^2/2} = 0$$ We can factor out \(e^{-x^2/2}\): $$e^{-x^2/2}(1-x^2) = 0$$ Since the exponential term is always positive, it can never be equal to zero. Therefore, we only need to focus on the term \((1-x^2)\): $$1 - x^2 = 0$$ Solve for x: $$x^2 = 1 \Rightarrow x = \pm 1$$ So, we have two critical points: \(x = 1\) and \(x = -1\).

Step 3: Determine intervals of increasing and decreasing

Now we need to analyze the first derivative \(f'(x)\) to find the intervals where the function is increasing or decreasing. We'll use the following test intervals: \((-\infty, -1)\), \((-1, 1)\), and \((1, \infty)\). 1. Test interval \((-\infty, -1)\): Choose a test point, such as \(x = -2\). Then \(f'(-2) = e^{-4/2} - (-2)^2e^{-4/2} = e^{-2}(1-4)=-3e^{-2}\). Since the first derivative at this test point is negative, the function is decreasing in this interval. 2. Test interval \((-1, 1)\): Choose a test point, such as \(x = 0\). Then \(f'(0) = e^{0} - 0^2e^{0} = e^{0}(1-0) = 1\). Since the first derivative at this test point is positive, the function is increasing in this interval. 3. Test interval \((1, \infty)\): Choose a test point, such as \(x = 2\). Then \(f'(2) = e^{-4/2} - 2^2e^{-4/2}= e^{-2}(1-4) = -3e^{-2}\). Since the first derivative at this test point is negative, the function is decreasing in this interval.

Step 4: Conclusion

Based on our analysis of the first derivative, the function \(f(x) = xe^{-x^2/2}\) is increasing on the interval \((-1, 1)\) and decreasing on the intervals \((-\infty, -1)\) and \((1, \infty)\).

Key Concepts

Derivative

Calculating the derivative of a function is a fundamental aspect of calculus. It helps us understand how a function changes and can indicate whether a function is increasing or decreasing at various points. To find the derivative of a function like \( f(x) = x e^{-x^2/2} \), we look for the rate at which \( f(x) \) changes with respect to \( x \). In simpler terms, it gives us a function, \( f'(x) \), that tells how the original function is evolving as \( x \) varies.

For practical purposes, like finding increasing or decreasing intervals, we often focus on the first derivative. If the first derivative \( f'(x) \) is positive over an interval, the function \( f(x) \) is increasing on that interval. Conversely, if \( f'(x) \) is negative, then \( f(x) \) is decreasing.

By applying derivative rules - such as the product rule and chain rule - we compute \( f'(x) \) for given functions. Understanding how to perform these calculations is key to plotting behaviors of complex functions.

Product Rule

The product rule is essential when differentiating functions that are the product of two or more distinct parts. For instance, in the function \( f(x) = x e^{-x^2/2} \), we have two separate parts: \( u = x \) and \( v = e^{-x^2/2} \). To differentiate such functions, the product rule states:

• \( (uv)' = u'v + uv' \)

To apply this, we first differentiate each part separately. Here, \( u = x \) gives us \( u' = 1 \), and \( v = e^{-x^2/2} \) is differentiated using the chain rule to give \( v' = -xe^{-x^2/2} \).

Plugging these results into the product rule formula, we find the derivative:

• \( f'(x) = 1 \, \cdot \, e^{-x^2/2} + x \, \cdot \, (-x)e^{-x^2/2} \)
• which simplifies to \( f'(x) = e^{-x^2/2} - x^2e^{-x^2/2} \)

The product rule thus provides a systematic approach to tackling derivatives of products, making it a powerful tool in calculus.

Critical Points

Critical points play a crucial role in understanding the behavior of functions. They are the points at which the first derivative of the function is zero or undefined. In other words, critical points are candidates for local maxima, minima, or points of inflection.

For our function \( f(x) = x e^{-x^2/2} \), finding the critical points involves setting the first derivative \( f'(x) = e^{-x^2/2} - x^2e^{-x^2/2} \) to zero. Solving \( f'(x) = 0 \), we factor out the non-zero term \( e^{-x^2/2} \) to get:

• \( e^{-x^2/2}(1-x^2) = 0 \)

Since the exponential term is never zero, we solve \( 1-x^2 = 0 \), thus yielding \( x = \pm 1 \) as critical points.

These points help determine intervals where the function is increasing or decreasing by testing the sign of \( f'(x) \) in regions around these points. If \( f'(x) \) changes sign at a critical point, it indicates a local maximum or minimum. Hence, understanding and finding critical points is crucial in graphing and analyzing functions' behavior.