Problem 67 - Thomas Calculus

Question

Find the absolute maximum and minimum values of \(f(x)=\) \(e^{x}-2 x\) on \([0,1]\).

Step-by-Step Solution

Answer

Absolute maximum is 1 at \(x = 0\); absolute minimum is \(2 - 2\ln(2)\).

Step 1: Identify Critical Points

First, we need to find the critical points of the function where the derivative is zero or undefined. The function is given by \(f(x) = e^x - 2x\). Calculate the derivative: \(f'(x) = e^x - 2\).Set \(f'(x) = 0\) to find critical points: \(e^x - 2 = 0\)=> \(e^x = 2\)=> \(x = \ln(2)\).

Step 2: Evaluate the Function at Critical Points and Endpoints

Now that we have the critical point \(x = \ln(2)\), and the endpoints \(x = 0\) and \(x = 1\), we will evaluate \(f(x)\) at these points.- At \(x = 0\): \(f(0) = e^0 - 2 \cdot 0 = 1\).- At \(x = 1\): \(f(1) = e^1 - 2 \cdot 1 = e - 2\).- At \(x = \ln(2)\): \(f(\ln(2)) = e^{\ln(2)} - 2 \cdot \ln(2) = 2 - 2\ln(2)\).

Step 3: Compare Values to Find Maximum and Minimum

We compare the function values at the critical point and the endpoints to determine the absolute maximum and minimum values within the interval.- \(f(0) = 1\)- \(f(1) = e - 2 \approx 0.718\)- \(f(\ln(2)) = 2 - 2\ln(2) \approx 0.613\)The highest value is \(f(0) = 1\), and the lowest value is \(f(\ln(2)) \approx 0.613\).

Key Concepts

Understanding Critical Points

Critical points are where a function's derivative equals zero or is undefined. These points are important because they might be where a function reaches a local maximum or minimum. To find these points, start by taking the derivative of the function. For our function, \( f(x) = e^x - 2x \), the derivative is \( f'(x) = e^x - 2 \).

• Set \( f'(x) \) to zero: \( e^x - 2 = 0 \)
• Solve for \( x \): \( e^x = 2 \) which gives \( x = \ln(2) \).

At \( x = \ln(2) \), the derivative is zero, indicating a critical point. In some scenarios, if the derivative is undefined, it may also result in a critical point, but that is not the case here. These points help us identify where the function changes direction, highlighting potential peaks or troughs within a defined interval.

Absolute Maxima and Minima

In calculus, absolute maxima and minima represent the highest and lowest points of a function on a given interval. Finding these points often involves examining the critical points and the endpoints of the interval. For the function \( f(x) = e^x - 2x \) on the interval \([0, 1]\), we evaluated the function at the critical point \( x = \ln(2) \) and the endpoints \( x = 0 \) and \( x = 1 \).

• At \( x = 0 \), \( f(0) = e^0 - 2 \times 0 = 1 \)
• At \( x = 1 \), \( f(1) = e^1 - 2 \times 1 = e - 2 \)
• At \( x = \ln(2) \), \( f(\ln(2)) = 2 - 2\ln(2) \)

After evaluating these points, the function's maximum value on \([0, 1]\) is \( f(0) = 1 \), and the minimum value is approximately \( f(\ln(2)) \approx 0.613 \). These points are unconditionally maximum and minimum within this interval.

Derivative Evaluation

The derivative of a function measures how the function's value changes as its input changes. It's a fundamental tool in calculus used for analyzing functions' behaviors. Evaluating derivatives help find critical points, detect function trends, and solve optimization problems.To evaluate the derivative of \( f(x) = e^x - 2x \):

• The derivative is found by applying basic rules of differentiation.
• For \( e^x \), the derivative is \( e^x \).
• For \(-2x\), the derivative is \(-2\).
• Combining these, we find \( f'(x) = e^x - 2 \).

This derivative tells us the rate of change of \( f(x) \) at any point \( x \). Setting \( f'(x) = 0 \) helps locate critical points and, ultimately, find where maxima and minima occur. This step is crucial because it identifies potential extreme values in the function's interval being analyzed.