Problem 119 - Thomas Calculus in SI Units

Question

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

Step-by-Step Solution

Answer

Maximum is 1 at \(x=0\), minimum is \(2-2\ln(2)\) at \(x=\ln(2)\).

Step 1: Identify the Critical Points

To find the critical points, we first need to calculate the derivative of the function, which is \( f'(x) = e^x - 2 \). Set this derivative equal to zero to find the critical points: \( e^x - 2 = 0 \). Solving this equation, we get \( e^x = 2 \), which implies \( x = \ln(2) \). Since \( x = \ln(2) \approx 0.693 \) lies within the interval \([0, 1]\), it is a critical point.

Step 2: Evaluate the Function at Critical Points and Endpoints

We evaluate \( f(x) \) at the critical point and at the endpoints of the interval \([0, 1]\):- At \( x = 0 \), \( f(0) = e^0 - 2 \times 0 = 1 \)- At \( x = 1 \), \( f(1) = e^1 - 2 \times 1 = e - 2 \)- At \( x = \ln(2) \), \( f(\ln(2)) = e^{\ln(2)} - 2\ln(2) = 2 - 2\ln(2) \)

Step 3: Determine the Absolute Maximum and Minimum

We compare the function values:- \( f(0) = 1 \)- \( f(1) = e - 2 \approx 0.718 \)- \( f(\ln(2)) = 2 - 2\ln(2) \approx 0.614 \)The absolute maximum value on \([0, 1]\) is 1 at \( x = 0 \). The absolute minimum value is \( 2 - 2 \ln(2) \) at \( x = \ln(2) \approx 0.693 \).

Key Concepts

Critical Points

In calculus, critical points are where the function’s derivative is either zero or undefined. These are significant because they are often points where the function reaches a local maximum or minimum. Finding critical points starts with taking the derivative of the function. In this exercise, we have the function \( f(x) = e^x - 2x \), and its derivative is \( f'(x) = e^x - 2 \).

To locate the critical points, we set \( f'(x) \) equal to zero and solve for \( x \). Here, \( e^x - 2 = 0 \) leads to \( e^x = 2 \), so \( x = \ln(2) \). It is important to check if this \( x \)-value is within the interval of interest, which it is in this case: \([0, 1]\). This means \( x = \ln(2) \) is indeed our critical point within this domain.

Critical points help us understand the function’s behavior and are crucial for finding extrema, including absolute maximum and minimum values of a function within a particular range.

Derivative

The derivative of a function provides us with essential information about its rate of change at any given point. It tells us how the function is increasing or decreasing and the slope of the tangent at any point on the curve.
To calculate the derivative of \( f(x) = e^x - 2x \), we differentiate each term. The exponential \( e^x \) remains \( e^x \) after differentiation, and the term \(-2x\) becomes \(-2\). Therefore, the derivative is \( f'(x) = e^x - 2 \).

Understanding derivatives is fundamental because:

• They help identify critical points by setting \( f'(x) = 0 \).
• They indicate where a function is increasing or decreasing by checking where \( f'(x) > 0 \) or \( f'(x) < 0 \).
• They provide insights into other phenomena like concavity and inflection points, though these are not our focus here.

With a derivative, we can gain insights into the function’s global and local behavior, making it easier to analyze and interpret real-world scenarios.

Absolute Maximum and Minimum

When we talk about absolute maximum and minimum, we are looking for the highest and lowest values that a function takes on an entire interval. For this, we check both critical points and interval endpoints.
To identify these values for \( f(x) = e^x - 2x \) on the interval \([0, 1]\), we evaluate the function at:

• The critical point \( x = \ln(2) \).
• The endpoints \( x = 0 \) and \( x = 1 \).

We calculated and compared these values:
\( f(0) = 1 \), \( f(1) = e - 2 \approx 0.718 \), and
\( f(\ln(2)) = 2 - 2 \ln(2) \approx 0.614 \).

Upon comparison, you see that the absolute maximum value is \( 1 \) at \( x = 0 \), while the absolute minimum value is \( 2 - 2 \ln(2) \) at \( x = \ln(2) \).
Determining these extreme values is crucial in many fields, from optimizing business strategies to understanding physical phenomena, as they tell us the limits of what is achievable or observeable.