Problem 54 - Single Variable Calculus

Question

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

Step-by-Step Solution

Answer

The function \(f(x) = e^{-x^2/2}\) is increasing in the interval \((- \infty, 0)\) and decreasing in the interval \((0, \infty)\).

Step 1: Find the first derivative

First, let's find the derivative of the function with respect to x. Using the chain rule, we obtain: \(f'(x) = \frac{d}{dx}(e^{-x^2/2}) = e^{-x^2/2} \cdot \frac{d}{dx}(-x^2/2) = e^{-x^2/2} \cdot (-x)\) So, \(f'(x) = -xe^{-x^2/2}\).

Step 2: Determine critical points

Next, let's identify the critical points of the function by finding the values of x for which \(f'(x) = 0\) or \(f'(x)\) doesn't exist: \(f'(x) = -xe^{-x^2/2}\) The function does exist for all the values of x, so let's set the derivative equal to zero and solve for x: \(-xe^{-x^2/2} = 0\) The only way for the product to be zero is if \(x = 0\). So, we have one critical point at \(x = 0\).

Step 3: Analyze the intervals

Now, we'll analyze the intervals, using the critical point to divide the domain of the function into intervals. We'll then determine whether the function is increasing or decreasing within each interval. Our intervals are \(-\infty \leq x \leq 0\) and \(0 \leq x \leq \infty\): 1. Interval \(x \in (-\infty, 0)\): We can pick a test point, say, \(x = -1\). \(f'(-1) = -(-1)e^{-(-1)^2/2} = e^{-1/2} > 0\). Since \(f'(x) > 0\), the function is increasing in this interval. 2. Interval \(x \in (0, \infty)\): We can pick a test point, say, \(x = 1\). \(f'(1) = -(1)e^{-(1)^2/2} = -e^{-1/2} < 0\). Since \(f'(x) < 0\), the function is decreasing in this interval.

Step 4: Conclusion

Based on our analysis, the function \(f(x) = e^{-x^2/2}\) is increasing in the interval \((- \infty, 0)\) and decreasing in the interval \((0, \infty)\).

Key Concepts

First Derivative Test

Understanding the First Derivative Test is like having a secret weapon for determining where a function's graph goes up or down. When you calculate the first derivative of a function, you're finding the slope at any point. If you imagine riding a bike along the graph, the first derivative tells you when you're pedaling uphill (increasing) or coasting downhill (decreasing).

To use the first derivative test, follow these steps:

• Find the first derivative of the function.
• Identify critical points where the derivative is zero or undefined.
• Choose test points around these critical points to determine the sign of the derivative in each interval.
• Then, if the derivative changes from positive to negative, you've found a local maximum. If it changes from negative to positive, you've got a local minimum!

Critical Points

The term Critical Points might sound like something out of a spy movie, but in math, it's where the action happens on a graph. Critical points occur where the derivative of a function is either zero or undefined. Think of them as checkpoints that tell you something important about the function's journey.

Why care about them?

• They help locate peaks (maximums) and valleys (minimums) of the function.
• They are key to sketching the graph of a function because they indicate where the graph changes direction.

To find them, set the derivative equal to zero and solve for the variable. Sometimes you might need to check where the derivative does not exist to find all the critical points. In our exercise, setting the derivative to zero gave us one critical point at \(x = 0\).

Chain Rule

The Chain Rule is like a magic spell for taking derivatives. It's used when you have a function nesting inside another function, like a Russian doll. Imagine you want to find the derivative of this nested situation; the chain rule tells you how to do it step by step.

Here's the chant:

• First, differentiate the outer function keeping the inner one unchanged.
• Then, multiply that by the derivative of the inner function.

When you apply this to our exercise, you start by taking the derivative of \(e^{stuff}\) as if 'stuff' is just \(x\), and then multiply by the derivative of the 'stuff' which is \(-x^2/2\). This is how the chain rule works its magic to get the derivative \(-xe^{-x^2/2}\).

Function Analysis

The big picture Function Analysis involves investigating all aspects of a function to understand its behavior. Here's what to look at:

• Domain and Range: Where the function lives and what values it can take.
• Intercepts: Where the function crosses the axes.
• Intervals of Increase/Decrease: Following the first derivative test to determine where the function goes up or down.
• Concavity and Inflection Points: Using the second derivative to find where the graph curves up or down.

When we applied function analysis to our exercise, we got the roadmap of \(f(x) = e^{-x^2/2}\) by using the first derivative test to find that the function ascends on the left side of the y-axis and descends on the right side.