Problem 34 - Calculus Single Variable

Question

Find the absolute minimum value and absolute maximum value of the given function on the given interval. \(f(x)=x^{2} e^{-x} ;[-2,3]\)

Step-by-Step Solution

Answer

The absolute minimum is 0 at \(x=0\) and the absolute maximum is \(4e^2\) at \(x=-2\).

Step 1: Determine Critical Points

To find the critical points, we need the derivative of the function. The given function is \( f(x) = x^2 e^{-x} \). Use the product rule to differentiate: \( f'(x) = \frac{d}{dx}(x^2) \cdot e^{-x} + x^2 \cdot \frac{d}{dx}(e^{-x}) = 2x e^{-x} - x^2 e^{-x} \). Simplify to \( f'(x) = e^{-x}(2x - x^2) \). Set \( f'(x) = 0 \) to find critical points: \( e^{-x}(2x - x^2) = 0 \), which simplifies to \( 2x - x^2 = 0 \). Solving gives \( x(x-2) = 0 \), so \( x = 0 \) or \( x = 2 \). Check if these points are in the interval \([-2, 3]\). Both are valid critical points.

Step 2: Evaluate Function at Critical Points

Evaluate the function \( f(x) \) at the critical points \( x = 0 \) and \( x = 2 \). Calculate \( f(0) = 0^2 e^0 = 0 \). Calculate \( f(2) = 2^2 e^{-2} = 4 e^{-2} \). This step finds the function values at the critical points.

Step 3: Evaluate Function at Endpoints of Interval

Evaluate the function \( f(x) \) at the endpoints of the interval \([-2, 3]\). Calculate \( f(-2) = (-2)^2 e^{2} = 4e^{2} \). Calculate \( f(3) = 3^2 e^{-3} = 9e^{-3} \). This gives us the function values at the interval boundaries.

Step 4: Compare Values to Find Absolute Extrema

Compare all calculated values: \( f(-2) = 4e^2 \), \( f(0) = 0 \), \( f(2) = 4e^{-2} \), \( f(3) = 9e^{-3} \). The smallest value is the absolute minimum, and the largest is the absolute maximum. Clearly, \( f(0) = 0 \) is the smallest value, making it the absolute minimum. The value \( f(-2) = 4e^2 \) is the largest, making it the absolute maximum.

Key Concepts

Critical Points

Critical points are key in determining the behavior of functions in calculus. These points occur where the derivative of a function is zero or undefined, signaling potential changes in direction or slope. To find critical points, we differentiate the function, so in this case, we differentiate the function \( f(x) = x^2 e^{-x} \) using the product rule, which results in \( f'(x) = e^{-x}(2x - x^2) \). This derivative is set to zero, \( e^{-x}(2x - x^2) = 0 \), which simplifies to \( 2x - x^2 = 0 \). Solving this gives us the critical points \( x = 0 \) and \( x = 2 \). Evaluating if these points lie within the given interval \( [-2, 3] \) confirms their relevance. Critical points help identify where function value extremes might occur.

Absolute Extrema

Finding the absolute extrema involves identifying the highest and lowest values of a function over a specified interval. This requires evaluating the function at critical points and at the interval's endpoints. The purpose is to determine the maximum and minimum values, known as the absolute extrema.
In solving for the function \( f(x) = x^2 e^{-x} \) over \( [-2, 3] \), we computed the function values at critical points \( x = 0 \) and \( x = 2 \), obtaining \( f(0) = 0 \) and \( f(2) = 4 e^{-2} \). Similarly, we checked the endpoints, \( f(-2) = 4e^2 \) and \( f(3) = 9e^{-3} \).
By comparing these values, the smallest value \( f(0) = 0 \) is identified as the absolute minimum, whereas the largest \( f(-2) = 4e^2 \) is the absolute maximum. Absolute extrema ensure the identification of these peaks over the interval considered.

Derivative

The derivative is a fundamental concept used to study functions' rates of change. It provides insights into where a function is increasing or decreasing and identifies critical points. Calculating a derivative requires mastery of rules like the product rule.
In the exercise, to find the derivative of \( f(x) = x^2 e^{-x} \), we applied the product rule: \( \frac{d}{dx}(u \, v) = u'v + uv' \). Here, \( u = x^2 \) and \( v = e^{-x} \), leading to \( f'(x) = 2x e^{-x} - x^2 e^{-x} = e^{-x}(2x - x^2) \).
The derivative is instrumental in identifying the critical points from which the extrema can be calculated. Understanding derivatives is crucial to solving calculus problems involving functions and their properties.

Interval Evaluation

Interval evaluation means assessing a function's behavior over a specified range of x-values. It includes checking boundaries and critical points within that interval.
In this exercise, we considered the interval \( [-2, 3] \), evaluating \( f(x) = x^2 e^{-x} \) at both the critical points and the endpoints. This method ensures that all potential extrema within the range are captured. \( f(-2) = 4e^2 \) and \( f(3) = 9e^{-3} \) were calculated at the interval's boundaries. Along with the critical points, these evaluations help identify the absolute extrema.

• Start by marking the interval boundaries.
• Compute function values at critical points.
• Assess function values at interval endpoints.

Through interval evaluation, comprehensive profiles of the function's behavior within the designated interval are revealed, simplifying the identification of absolute maxima and minima.