Problem 41 - University Calculus: Early Transcendentals

Question

a. Find the open intervals on which the function is increasing and decreasing. b. Identify the function's local and absolute extreme values, if any, saying where they occur. $$f(x)=e^{2 x}+e^{-x}$$

Step-by-Step Solution

Answer

It decreases on \((-\infty, \frac{\ln(\frac{1}{2})}{3}]\) and increases on \((\frac{\ln(\frac{1}{2})}{3}, \infty)\), with a local minimum at \(x = \frac{\ln(\frac{1}{2})}{3}\).

Step 1: Find the Derivative

To determine intervals where the function is increasing or decreasing, we must first find the derivative. For the function \(f(x) = e^{2x} + e^{-x}\), the derivative \(f'(x)\) is found as follows:\[ f'(x) = \frac{d}{dx}(e^{2x}) + \frac{d}{dx}(e^{-x}) = 2e^{2x} - e^{-x}. \]

Step 2: Find Critical Points

To find the critical points, set \(f'(x) = 0\):\[ 2e^{2x} - e^{-x} = 0. \]Multiply through by \(e^x\) to clear the exponent:\[ 2e^{3x} = 1. \]Solve for \(x\) by taking the natural logarithm:\[ e^{3x} = \frac{1}{2} \]\[ 3x = \ln\left(\frac{1}{2}\right) \]\[ x = \frac{\ln\left(\frac{1}{2}\right)}{3}. \]

Step 3: Analyze Critical Points

Evaluate the sign of \(f'(x)\) around the critical point:- Choose a test point less than \(x = \frac{\ln(\frac{1}{2})}{3}\), such as \(x = -1\). Calculate \(f'(-1) = 2e^{-2} - e = 2\cdot(1/e^2) - e\); this is negative as e dominates.- Choose a test point greater than \(x = \frac{\ln(\frac{1}{2})}{3}\), such as \(x = 0\). Calculate \(f'(0) = 2e^0 - e^0 = 2 - 1 = 1\); this is positive.The function is decreasing on \((-\infty, \frac{\ln(\frac{1}{2})}{3}]\) and increasing on \((\frac{\ln(\frac{1}{2})}{3}, \infty)\).

Step 4: Determine Local and Absolute Extrema

Since the function transitions from decreasing to increasing at \(x = \frac{\ln(\frac{1}{2})}{3}\), it has a local minimum there.Substitute \(x = \frac{\ln(\frac{1}{2})}{3}\) into \(f(x)\) to find the local minimum:\[ f\left(\frac{\ln(\frac{1}{2})}{3}\right) = e^{2 \left(\frac{\ln(\frac{1}{2})}{3}\right)} + e^{- \left(\frac{\ln(\frac{1}{2})}{3}\right)}. \]

Key Concepts

Derivative

The derivative of a function is an essential tool in calculus. It's like a mathematical gadget that measures how a function changes at any point. In simple words, the derivative tells us the rate of change or the slope of the function at a certain point. For the function \(f(x) = e^{2x} + e^{-x}\), the process of finding the derivative involves using basic differentiation rules for exponential functions. You compute the derivative \(f'(x)\) by differentiating each term separately:

• For \(e^{2x}\), apply the chain rule to get \(2e^{2x}\).
• For \(e^{-x}\), take the derivative to get \(-e^{-x}\).

Combining these results gives us the derivative: \(f'(x) = 2e^{2x} - e^{-x}\). Understanding the derivative is the first step in analyzing the behavior of a function.

Critical Points

Critical points are where the function's rate of change is zero or undefined. These points are crucial as they often indicate peaks, valleys, or flat spots on the graph. To find the critical points of a function like \(f(x) = e^{2x} + e^{-x}\), set the derivative \(f'(x) = 2e^{2x} - e^{-x}\) to zero and solve for \(x\). This involves:

• Setting \(2e^{2x} - e^{-x} = 0\).
• Multiplying through by \(e^x\) to eliminate fractions, giving \(2e^{3x} = 1\).
• Taking the natural log, resulting in \(3x = \ln\left(\frac{1}{2}\right)\). Solve for \(x\) to find \(x = \frac{\ln\left(\frac{1}{2}\right)}{3}\).

These steps reveal the critical point, a potential location for local extrema.

Local Extrema

Local extrema refer to the peaks (local maximum) or valleys (local minimum) of a function within a specified interval. After finding the critical points, investigate if they correspond to local maxima or minima. Using the derivative sign analysis helps here:

• We tested points around the critical point \(x = \frac{\ln(\frac{1}{2})}{3}\).
• Found that the function changes from decreasing on \((\infty, \frac{\ln(\frac{1}{2})}{3}]\) to increasing on \((\frac{\ln(\frac{1}{2})}{3}, \infty)\).

Such behavior indicates a local minimum at the critical point. By substituting \(x = \frac{\ln(\frac{1}{2})}{3}\) back into the original function \(f(x)\), you find the actual value at the local minimum.

Increasing and Decreasing Functions

To understand where a function is increasing or decreasing, look at the sign of the derivative. The function increases where the derivative \(f'(x)\) is positive and decreases where it's negative.

• For the function \(f(x)=e^{2x}+e^{-x}\), we determined that \(f'(x)=2e^{2x}-e^{-x}\).
• The sign of \(f'(x)\) was tested around the critical point \(x = \frac{\ln(\frac{1}{2})}{3}\):
• At \(x = -1\), \(f'(-1)\) was negative, showing the function is decreasing before the critical point.
• At \(x = 0\), \(f'(0)\) turned positive, indicating the function is increasing after the critical point.

Thus, we conclude the intervals where the function rises or falls, crucial for sketching or understanding the graph's behavior. Properly identifying these intervals helps in comprehending the function's trends and structure.