Problem 40 - Calculus Early Transcendentals

Question

Use a graphing utility to estimate the absolute maximum and minimum values of \(f \), if any, on the stated interval, and then use calculus methods to find the exact values. \(f(x)=\left(x^{2}-1\right) e^{x} ;[-2,2]\)

Step-by-Step Solution

Answer

Absolute minimum: approximately \(0.406\) at \(x = -2\); Absolute maximum: approximately \(22.167\) at \(x = 2\).

Step 1: Graphing the Function

Start by graphing the function \[ f(x) = (x^2 - 1)e^x \] over the interval \([-2, 2]\) using a graphing calculator or software. Analyze the graph to identify the approximate location of the absolute minimum and maximum values.

Step 2: Finding Critical Points

To find the exact maximum and minimum values, first identify the critical points where the first derivative is zero or undefined. Calculate the first derivative:\[ f'(x) = e^x(2x) + (x^2 - 1)e^x = (x^2 + 2x - 1)e^x. \]Set the derivative equal to zero to find critical points:\[ x^2 + 2x - 1 = 0. \]

Step 3: Solving the Quadratic Equation

Solve the quadratic equation \[ x^2 + 2x - 1 = 0 \] usign the quadratic formula \[ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}, \] where \( a = 1, b = 2, c = -1 \). This gives\[ x = \frac{-2 \pm \sqrt{4 + 4}}{2} = \frac{-2 \pm 2\sqrt{2}}{2}, \] resulting in the roots \( x = -1 + \sqrt{2} \) and \( x = -1 - \sqrt{2} \).

Step 4: Evaluating at Critical Points and Endpoints

Evaluate the function \[ f(x) = (x^2 - 1)e^x \] at the critical points \( x = -1 + \sqrt{2} \) and \( x = -1 - \sqrt{2} \), as well as the endpoints of the interval, \( x = -2 \) and \( x = 2 \). These evaluations provide potential extrema:- For \( x = -2 \), calculate \( f(-2) = ((-2)^2 - 1)e^{-2} = 3e^{-2} \).- For \( x = 2 \), calculate \( f(2) = (2^2 - 1)e^2 = 3e^2 \).- For \( x = -1 + \sqrt{2} \), calculate \( f(-1 + \sqrt{2}) \).- For \( x = -1 - \sqrt{2} \), calculate \( f(-1 - \sqrt{2}) \).

Step 5: Comparing Values

Compare the values obtained from the evaluations:- \( f(-2) = 3e^{-2} \approx 0.406 \)- \( f(2) = 3e^2 \approx 22.167 \)- Calculate numerical values for \( f(-1 + \sqrt{2}) \) and \( f(-1 - \sqrt{2}) \) using a calculator.From these comparisons, deduce the absolute maximum and minimum on the interval \([-2, 2]\).

Key Concepts

Absolute Maximum

In calculus, the absolute maximum of a function is the highest point over a specified interval. To determine this, one must evaluate the function at all critical points and endpoints within the given interval. In our example function,

• we use the interval given as to find the absolute maximum.

By calculating the function values at the critical points and endpoints, we discovered that the absolute maximum is at the point where .

1. The function reaches its peak value of .
2. It demonstrates the highest point the curve attains on this interval.
   This is crucial in various applications such as economics and engineering where finding maximum output or efficiency is desired.

Absolute Minimum

The absolute minimum represents the lowest value a function achieves on a given interval. Like finding the maximum, one must check the function's behavior at critical points and endpoints within the interval. For we calculated this across the interval

• The function value was lowest at , which was .
• This marks the absolute minimum point on the graph.

Recognizing the absolute minimum is vital for understanding constraints and limitations within a system, such as cost minimization or minimum required resources in practical scenarios.

Critical Points

Critical points of a function occur where its derivative is zero or undefined. Finding these points helps to identify where potential maxima, minima, or other interesting features like plateaus might occur. For , we computed the first derivative
Setting this equal to zero revealed the quadratic equation .

• Solving this equation provided us with critical points at and .
• These are the locations where the slope of the tangent to the curve is zero, indicating potential extremum points or flat regions.

Critical points are especially useful in detailed analysis of functions to predict and control behavior in scenarios from logistics to physics.

Graphing Utility

A graphing utility is an essential tool in calculus for visualizing a function's behavior over a certain interval.

• By graphically representing , you can estimate where the highest and lowest points might be.
• These utilities provide a visual reference, making it easier to understand how a function behaves between defined endpoints.

In our exercise, using such a tool helped us initially pinpoint the approximate locations of the absolute maximum and minimum. This aids not only in confirming the results obtained by calculus but also in providing an intuitive grasp of the function's overall trajectory. Graphing utilities, like software or calculators, are invaluable in educational settings, offering robust support in exploring and solving complex mathematical problems.