Problem 33 - Calculus Single Variable

Question

Find the absolute minimum value and absolute maximum value of the given function on the given interval. \(f(x)=e^{x}-x ;[-1,1]\)

Step-by-Step Solution

Answer

The absolute minimum is approximately 0.368 and the absolute maximum is approximately 1.718.

Step 1: Find the Derivative

To find critical points, we first need the derivative of the function. The derivative of the function \(f(x) = e^x - x\) is \(f'(x) = e^x - 1\).

Step 2: Solve for Critical Points

Set the derivative equal to zero to find the critical points: \(e^x - 1 = 0\). Solving this gives \(e^x = 1\). So, \(x = \ln(1) = 0\).

Step 3: Evaluate the Function at Endpoints

Now evaluate the function at the endpoints of the interval. For \(x = -1\), \(f(-1) = e^{-1} + 1 \approx 0.368\). For \(x = 1\), \(f(1) = e^1 - 1 = e - 1 \approx 1.718\).

Step 4: Evaluate the Function at Critical Points

Evaluate the function at the critical point inside the interval. For \(x = 0\), \(f(0) = e^0 - 0 = 1\).

Step 5: Determine Absolute Extremes

Compare the values of \(f(x)\) at the endpoints and the critical point: \(f(-1) \approx 0.368\), \(f(0) = 1\), \(f(1) \approx 1.718\). The absolute minimum is \(f(-1) \approx 0.368\) and the absolute maximum is \(f(1) \approx 1.718\).

Key Concepts

Derivative calculation

The first step in solving optimization problems in calculus is to calculate the derivative of the function. Derivatives help us understand how a function behaves by showing the rate at which it changes. In our example, the function is given as \(f(x) = e^x - x\). Calculating the first derivative gives us \(f'(x) = e^x - 1\).

• The derivative \(e^x\) represents the rate of change of the exponential function \(e^x\), which grows rapidly.
• The \(-1\) is the derivative of \(-x\), representing a constant rate of decline.

This derivative tells us where the function is increasing or decreasing, which is essential for locating critical points, the next step in finding extreme values.

Critical points

Critical points are where the function's rate of change either stops or changes direction. They occur where the first derivative is equal to zero or does not exist.
In the exercise, we set the derivative equal to zero to find the critical points: \(e^x - 1 = 0\).

• Solving \(e^x = 1\) gives \(x = \ln(1) = 0\).
• This means that the function has one critical point at \(x = 0\), within the interval \([-1,1]\).

Finding these critical points allows us to further investigate if these points might yield a maximum or a minimum value of the function.

Function evaluation

Once the critical points are identified, we must evaluate the function at these points and at the endpoints of the interval. This helps us determine the function's behavior over the entire interval.
For our function \(f(x) = e^x - x\), we evaluate at:

• Endpoint \(x = -1\), giving \(f(-1) = e^{-1} + 1 \approx 0.368\).
• Endpoint \(x = 1\), resulting in \(f(1) = e - 1 \approx 1.718\).
• Critical point \(x = 0\), where \(f(0) = 1\).

Evaluating the function at these points ensures a comprehensive check of potential extreme values.

Absolute extrema

Absolute extrema refer to the highest and lowest values a function can take on a given domain. To find the absolute minimum and maximum values within an interval, we compare the function's values at critical points and endpoints.
From our evaluations, we have:

• \(f(-1) \approx 0.368\)
• \(f(0) = 1\)
• \(f(1) \approx 1.718\)

Based on this, the absolute minimum value is \(f(-1) \approx 0.368\) and the absolute maximum value is \(f(1) \approx 1.718\).
Understanding absolute extrema helps us fully grasp how the function behaves across the entire interval and identifies the most significant points of interest.