Problem 38 - Applied Mathematics: For the Managerial, Life, and Social Sciences

Question

Find the absolute maximum value and the absolute minimum value, if any, of each function. $$ f(x)=x e^{-x^{2}} \text { on }[0,2] $$

Step-by-Step Solution

Answer

The absolute maximum value of the function \(f(x)=x e^{-x^{2}}\) on [0, 2] is approximately 0.4289 at \(x=\frac{1}{\sqrt{2}}\), and the absolute minimum value is 0 at x=0.

Step 1: Find Critical Points

To find the critical points, we need to find the first derivative of the function. Given the function \(f(x) = xe^{-x^2}\), we can use the product rule for differentiation: \((uv)' = u'v + uv'\) Here, \(u = x\) and \(v = e^{-x^2}\) \begin{align*} u' &= \frac{d}{dx}(x) = 1 \\ v' &= \frac{d}{dx}(e^{-x^2}) = e^{-x^2} (-2x) \end{align*} Using the product rule, we compute the first derivative: \[ f'(x) = 1 \cdot e^{-x^2} + x \cdot (-2x) e^{-x^2} = e^{-x^2}(1 - 2x^2) \] Now, we set \(f'(x)\) equal to zero and solve for x: \[ e^{-x^2}(1 - 2x^2) = 0 \]

Step 2: Solve for Critical Points

Since \(e^{-x^2}\) is never equal to 0, the equation simplifies to: \[ 1 - 2x^2 = 0 \] Rearranging and solving for x: \[ 2x^2 = 1 \] \[ x^2 = \frac{1}{2} \] \[ x = \pm\frac{1}{\sqrt{2}} \] However, only \(x=\frac{1}{\sqrt{2}}\) is in the interval [0, 2]. So, we have one critical point: \(x=\frac{1}{\sqrt{2}}\).

Step 3: Check Endpoints and Compare Function Values

Now we need to evaluate the function at the critical point and the given endpoints to find the absolute maximum and minimum values: \[ f\left(\frac{1}{\sqrt{2}}\right) = \frac{1}{\sqrt{2}}e^{-\left(\frac{1}{2}\right)} \\ f(0) = (0)e^{-0^2} = 0 \\ f(2) = (2)e^{-2^2} = 2e^{-4} \] Now we compare the function values: \[ f\left(\frac{1}{\sqrt{2}}\right) \approx 0.4289, f(0) = 0, f(2) \approx 0.0366 \] From the comparison, we can see that the absolute maximum value is approximately 0.4289 at \(x=\frac{1}{\sqrt{2}}\) and the absolute minimum value is 0 at x=0.

Key Concepts

Critical Points

Critical points are fundamental in calculus as they help identify where a function's slope changes, potentially marking maxima, minima, or points of inflection. To find critical points, you start by computing the first derivative of the function. This derivative represents the slope. The critical points occur where this derivative is zero or undefined. In the given function, \(f(x) = xe^{-x^2}\), finding the derivative involves using the product rule.

• Product Rule: If you have two functions multiplied together, \(u\) and \(v\), the derivative is \((uv)' = u'v + uv'\).
• For \(f(x)\), apply this rule with \(u = x\) and \(v = e^{-x^2}\).
• The resulting derivative is \(f'(x) = e^{-x^2}(1 - 2x^2)\).

Once you find the derivative, set it equal to zero to solve for the critical points. This solution indicates where the slope is zero, a necessary condition for identifying potential local maximum or minimum points. In this example, within the interval \([0,2]\), the critical point is \(x = \frac{1}{\sqrt{2}}\).

Derivative

The derivative is a core concept in calculus, representing the rate of change of a function at any given point. Calculating the derivative helps to find where a function is increasing or decreasing, hence highlighting critical points.

• In practice, this is the slope of the tangent line to the curve at any point \(x\).
• For our function \(f(x) = xe^{-x^2}\), the product rule was applied to get \(f'(x) = e^{-x^2}(1 - 2x^2)\).
• This derivative incorporates both the differentiation of the function \(x\) and the exponential function \(e^{-x^2}\), showcasing the use of multiple differentiation rules in a single calculation.

Understanding how to derive these functions and interpret their derivatives provides insight into the behavior of the function itself. It is essential for tasks like finding maximum/minimum values, analyzing motion, and understanding changes over intervals.

Absolute Maximum and Minimum

The absolute maximum and minimum are the highest and lowest values a function takes within a specified interval. They may occur at endpoints of the interval or at critical points within it. To find these extreme values:

• First, determine critical points by solving where the derivative equals zero.
• Second, evaluate the function at these critical points and at the endpoints of your interval.
• Finally, compare these values to identify the highest and lowest outputs.

In our exercise, the interval \([0, 2]\) contains one critical point at \(x = \frac{1}{\sqrt{2}}\). Evaluating the function at this point and at \(x = 0\) and \(x = 2\), we find:- \(f(0) = 0\) is the absolute minimum.- \(f\left(\frac{1}{\sqrt{2}}\right) \approx 0.4289\) is the absolute maximum.- \(f(2) \approx 0.0366\) is lower than \(f\left(\frac{1}{\sqrt{2}}\right)\) making it neither the max nor min within \([0, 2]\).

Interval Evaluation

Interval evaluation is the process of checking a function's behavior over a specific range of values. It is critical when you need to find absolute extremum points. When given an interval:

• Always check the value of the function at the endpoints extra carefully because absolute extrema may occur there.
• Consider how the function behaves in relation to its critical points within the interval.
• Evaluate all potential points where the function could achieve significant highs or lows.

For the function \(f(x) = xe^{-x^2}\), we carefully evaluate \(f(x)\) at critical point \(x=\frac{1}{\sqrt{2}}\), as well as endpoints \(x = 0\) and \(x = 2\). The careful evaluation solidifies understanding of where the extremes are located, ensuring none are overlooked, which is a fundamental practice in calculus when comparing potential maximum and minimum values over any given interval.