Problem 23 - Calculus Early Transcendentals: Pearson New International Edition

Question

Identify the critical points and find the maximum value and minimum value on the given interval. $$ f(x)=x e^{-x^{2}} ; I=[-1,2] $$

Step-by-Step Solution

Answer

The critical points are \( x = \pm \frac{1}{\sqrt{2}} \); max: \( \frac{1}{\sqrt{2}} e^{-\frac{1}{2}} \), min: \( -\frac{1}{\sqrt{2}} e^{-\frac{1}{2}} \).

Step 1: Find the derivative

To find critical points, we first need to find the derivative of the function. Given the function \( f(x) = x e^{-x^2} \), we apply the product rule: \( f'(x) = e^{-x^2} + x(-2x e^{-x^2}) = e^{-x^2} (1 - 2x^2) \).

Step 2: Solve for critical points

Set the derivative equal to zero to find critical points:\[ e^{-x^2} (1 - 2x^2) = 0 \]Since \( e^{-x^2} \) is never zero, we solve:\[ 1 - 2x^2 = 0 \]\[ 2x^2 = 1 \]\[ x^2 = \frac{1}{2} \]\[ x = \pm \frac{1}{\sqrt{2}} \] These are the critical points: \( x = \frac{1}{\sqrt{2}} \) and \( x = -\frac{1}{\sqrt{2}} \).

Step 3: Evaluate the function at critical points and endpoints

We need to evaluate \( f(x) \) at the critical points and the endpoints of the interval \([-1, 2]\).- Evaluate at \( x = -1 \): \[ f(-1) = -1 e^{-(-1)^2} = -1 e^{-1} \]- Evaluate at \( x = 2 \): \[ f(2) = 2 e^{-4} \]- Evaluate at \( x = \frac{1}{\sqrt{2}} \): \[ f\left( \frac{1}{\sqrt{2}} \right) = \frac{1}{\sqrt{2}} e^{-\frac{1}{2}} \]- Evaluate at \( x = -\frac{1}{\sqrt{2}} \): \[ f\left( -\frac{1}{\sqrt{2}} \right) = -\frac{1}{\sqrt{2}} e^{-\frac{1}{2}} \]

Step 4: Determine maximum and minimum values

Compare the values obtained:- \( f(-1) = -\frac{1}{e} \)- \( f(2) = 2 e^{-4} \)- \( f\left( \frac{1}{\sqrt{2}} \right) = \frac{1}{\sqrt{2}} e^{-\frac{1}{2}} \)- \( f\left( -\frac{1}{\sqrt{2}} \right) = -\frac{1}{\sqrt{2}} e^{-\frac{1}{2}} \)Among these, \( \frac{1}{\sqrt{2}} e^{-\frac{1}{2}} \) is largest, hence the maximum value, and \( -\frac{1}{\sqrt{2}} e^{-\frac{1}{2}} \) is smallest, hence the minimum value.

Key Concepts

Critical Points

Critical points are where the behavior of a function can change, such as transitioning from increasing to decreasing. For any differentiable function, these are the points where the derivative equals zero or doesn't exist. Here's how you can identify them:

• Take the derivative of the function.
• Set the derivative equal to zero.
• Solve the equation to find critical values where the function's rate of change is zero.

In our exercise, after finding the derivative of the function \( f(x) = x e^{-x^2} \), we set this equal to zero to find \( x = \pm \frac{1}{\sqrt{2}} \). This shows us the locations where the function might have a local extremum (a peak or valley) or a point of inflection.

Maximum and Minimum Values

Maximum and minimum values are the highest and lowest points a function attains on a given interval. They can be found among the critical points and the endpoints of the interval. Here's how it's done:

• Evaluate the function at each of the critical points.
• Evaluate the function at the endpoints of the interval.
• Compare these function values to determine the largest (maximum) and smallest (minimum) values.

For the function \( f(x) = x e^{-x^2} \) over \( I=[-1,2] \), we calculated and compared the values of \( f(x) \) at the critical points and endpoints, concluding that the maximum value is \( \frac{1}{\sqrt{2}} e^{-\frac{1}{2}} \) and the minimum value is \( -\frac{1}{\sqrt{2}} e^{-\frac{1}{2}} \).

Derivative

The derivative of a function gives us the function's rate of change. Understanding it is crucial for finding critical points. To find the derivative:

• Use rules of differentiation, such as the product rule or chain rule, as needed.
• Simplify the expression to make further calculations easier.

In our example, the function \( f(x) = x e^{-x^2} \) is handled by using the product rule to find \( f'(x) = e^{-x^2} (1 - 2x^2) \). This derivative tells us how \( f(x) \) behaves at each point, helping us locate where it could have peaks or troughs.

Function Evaluation

Function evaluation involves plugging specific values into a function to calculate the output. This process is vital when determining where a function reaches its maximum and minimum over an interval. The steps include:

• Substitute the critical points and interval endpoints into the function.
• Calculate the function values to see which are higher or lower.

For our function \( f(x) = x e^{-x^2} \), we carefully evaluated the function at the critical points \( x = \pm \frac{1}{\sqrt{2}} \) and at the endpoints \( x = -1, 2 \). This allowed us to identify both the maximum and minimum values effectively.