Problem 58 - Single Variable Calculus

Question

In Exercises \(41-60\), find the absolute maximum and absolute minimum values, if any, of the function. $$ f(x)=e^{2 x}-e^{x} ; \quad[-2,0] $$

Step-by-Step Solution

Answer

The absolute minimum value of the function \(f(x) = e^{2x} - e^x\) on the interval \([-2, 0]\) is \(f(-2) = e^{-4} - e^{-2}\), and the absolute maximum value is \(f(0) = 0\).

Step 1: Calculate the first derivative of the function

To find the critical points of the function, we must first compute its first derivative: \[ f'(x) = \frac{d}{dx}(e^{2x} - e^x) = 2e^{2x} - e^x \]

Step 2: Find the critical points of the function

Now, we will set the first derivative equal to zero and solve for x: \[ 2e^{2x} - e^x = 0 \] To solve this equation, we can factor out \(e^x\): \[ e^x (2e^x - 1) = 0 \] Thus, we can find the critical points by setting each factor to zero: \[ e^x = 0 \quad \text{or} \quad 2e^x - 1 = 0 \] Since \(e^x\) is never zero, there are no critical points for the first equation. For the second equation, we can solve for x: \[ 2e^x = 1 \Rightarrow e^x = \frac{1}{2} \Rightarrow x = \ln\left(\frac{1}{2}\right) = -\ln 2 \] Since \(-2 \leq -\ln2 \leq 0\), it is a critical point within the interval.

Step 3: Evaluate the function at the critical point and interval's endpoints

We will now evaluate the function at the critical point found in Step 2 and at the interval's endpoints: \(-2, -\ln2, 0\): \[ f(-2) = e^{2(-2)} - e^{-2} = e^{-4} - e^{-2} \] \[ f(-\ln2) = e^{2(-\ln2)} - e^{-\ln2} = e^{-2\ln2} - e^{-\ln2} = \frac{1}{4} - \frac{1}{2} \] \[ f(0) = e^{2(0)} - e^{0} = 1 - 1 = 0 \]

Step 4: Identify the absolute maximum and minimum values

Now, we will compare the function values from Step 3 to determine the absolute maximum and minimum values: \[ f(-2) < f(-\ln2) < f(0) \] Thus, the absolute minimum value is \(f(-2) = e^{-4} - e^{-2}\) and the absolute maximum value is \(f(0) = 0\).

Key Concepts

First Derivative

When trying to find the absolute maximum and minimum values of a function, the first step is to compute its first derivative. This step is crucial because the first derivative can reveal important information about the function's behavior. Essentially, the first derivative, denoted as \( f'(x) \), represents the rate of change of the function \( f(x) \). For the given function, \( f(x) = e^{2x} - e^x \), the first derivative is found using the rules of differentiation:

• The derivative of \( e^{2x} \) is \( 2e^{2x} \) due to the chain rule.
• The derivative of \( e^x \) is simply \( e^x \) itself.

Therefore, the first derivative is \( f'(x) = 2e^{2x} - e^x \). This derivative provides the slope of the tangent line to the function at any point \( x \) and helps us find critical points.

Critical Points

Critical points are x-values where the behavior of the function drastically changes, which can be found by setting the first derivative equal to zero and solving for \( x \). For our function, we solve \( 2e^{2x} - e^x = 0 \) by factoring:

• Factor out \( e^x \): \( e^x(2e^x - 1) = 0 \).
• Since \( e^x \) is always positive, it can never be zero, so we focus on the term \( 2e^x - 1 = 0 \).
• Solving \( 2e^x = 1 \) gives \( e^x = \frac{1}{2} \), resulting in \( x = -\ln(2) \).

This \( x = -\ln(2) \) is the critical point we seek. Critical points are pivotal in determining where a function has potential maximum or minimum values within a given interval.

Function Evaluation

Once critical points are identified, the next step is to evaluate the function at both these points and the endpoints of the interval where the function is defined. For our interval, \([-2, 0]\), we perform function evaluations at \( x = -2, -\ln(2),\) and \( 0 \):

• Calculating \( f(-2) = e^{-4} - e^{-2} \).
• For \( x = -\ln 2 \), evaluate \( f(-\ln 2) = \frac{1}{4} - \frac{1}{2} \).
• Lastly, at \( x = 0 \), we find \( f(0) = 1 - 1 = 0 \).

These evaluations help us compare the function's values, thereby helping us identify where absolute maximum and minimum values occur. Evaluating the function at the critical points provides possible candidates for the function's extrema on the interval.

Interval Analysis

Interval analysis involves comparing the values obtained from evaluating the function at the critical points and interval endpoints to determine the absolute maximum and minimum values of the function on a given interval. For \( f(x) = e^{2x} - e^x \) over the interval \([-2, 0]\), we compare:

• The function value at \( x = -2 \), which is \( e^{-4} - e^{-2} \).
• The value at the critical point \( x = -\ln 2 \), which is \( \frac{1}{4} - \frac{1}{2} \).
• The value at \( x = 0 \), which is \( 0 \).

Through comparison, \( f(0) = 0 \) is identified as the absolute maximum, while \( f(-2) = e^{-4} - e^{-2} \) is identified as the absolute minimum. Proper interval analysis ensures that we accurately determine the global extrema of the function within the specified interval.