Problem 23 - Calculus with Concepts in Calculus

Question

Use the First Derivative Test to determine the relative extreme values (if any) of the function. $$ f(x)=x^{2} e^{-x} $$

Step-by-Step Solution

Answer

The function has a relative maximum at \( x = 2 \).

Step 1: Find the first derivative

First, we need to find the derivative of the given function \( f(x) = x^{2}e^{-x} \). We will use the product rule, which states that \((uv)' = u'v + uv'\), for \( u = x^2 \) and \( v = e^{-x} \). We also use the derivative of \( e^{-x} \), which is \(-e^{-x}\). The first derivative is: \[ f'(x) = 2xe^{-x} - x^{2}e^{-x} = e^{-x}(2x - x^{2}) \].

Step 2: Find critical points

To find critical points, we set the first derivative equal to zero and solve for \( x \): \[ e^{-x}(2x - x^{2}) = 0 \]. Since \( e^{-x} \) is never zero, solve \( 2x - x^{2} = 0 \), which simplifies to \( x(x - 2) = 0 \). Thus, critical points are \( x = 0 \) and \( x = 2 \).

Step 3: Apply the First Derivative Test

Now, we apply the First Derivative Test to determine the nature of each critical point. We check the sign change of \( f'(x) \) around each critical point. - **Interval for \( x = 0 \)**: Choose test points, like \( x = -1 \) and \( x = 1 \). - \( f'(-1) = e^{1}(2(-1) - (-1)^2) = e(2 + 1) > 0 \), so \( f'(x) > 0 \) for \( x < 0 \). - \( f'(1) = e^{-1}(2(1) - 1^2) = e^{-1}(1) > 0 \), so \( f'(x) > 0 \) for \( 0 < x < 2 \). - **Interval for \( x = 2 \)**: Choose test points, like \( x = 1.5 \) and \( x = 3 \). - \( f'(1.5) = e^{-1.5}(2(1.5) - (1.5)^2) = e^{-1.5}(3 - 2.25) > 0 \), so \( f'(x) > 0 \) for \( x < 2 \). - \( f'(3) = e^{-3}(2(3) - 3^2) = e^{-3}(6 - 9) < 0 \), which means \( f'(x) < 0 \) for \( x > 2 \).

Key Concepts

Critical Points

Critical points of a function are values of \( x \) where the derivative is either zero or undefined.
These points are important because they are potential locations for relative extrema, which are the highest or lowest points in a certain interval on the graph.
To find critical points, we follow these steps:

• Compute the first derivative of the function.
• Set the derivative equation to zero and solve for \( x \).
• Check whether any values of \( x \) make the derivative undefined.

In our example, we found that the critical points of the function \( f(x) = x^2e^{-x} \) are \( x = 0 \) and \( x = 2 \).
These points are derived from setting \( f'(x) = e^{-x}(2x - x^2) \) to zero and solving \( 2x - x^2 = 0 \).
This equation factors to \( x(x - 2) = 0 \). Thus, the solutions \( x = 0 \) and \( x = 2 \) are our critical points.

Derivative of a Function

The derivative of a function is a core concept in calculus.
It measures how the function value changes as the input changes, essentially giving the slope of the tangent line to the graph of the function at any point.
Calculating a derivative involves distinguishing various rules and applying them appropriately, such as the product rule and the chain rule.

• The derivative tells us whether the function is increasing, decreasing, or has any flat points (i.e., zero slope).
• If the derivative is positive over an interval, the function is increasing in that interval.
• If the derivative is negative, the function is decreasing.
• If the derivative is zero, it may indicate a flat (horizontal) point.

In the function \( f(x) = x^2e^{-x} \), its derivative is \( f'(x) = e^{-x}(2x - x^2) \).
Here, we look at where \( f'(x) \) is positive or negative to understand the behavior of the original function.

Product Rule

The product rule is a fundamental rule in calculus used when differentiating products of two functions.
It states that if \( u \) and \( v \) are both functions of \( x \), then the derivative of their product is given by: \((uv)' = u'v + uv'\).
This rule helps us find derivatives of more complex functions, much like our example function.

• Identify the two parts of your function that are being multiplied.
• Differentiate each part separately.
• Apply the product rule formula.

In the given function \( f(x) = x^2 e^{-x} \), we identify \( u = x^2 \) and \( v = e^{-x} \).
Using the product rule, we find the derivative as: \( f'(x) = 2x \, e^{-x} - x^2 \, e^{-x} = e^{-x}(2x - x^2) \).
This application highlights the power of the product rule in simplifying the process of differentiation.

Relative Extrema

Relative extrema refer to the local maximum or minimum values that a function can attain within a given interval.
They aren't necessarily the highest or lowest values of an entire function, just within that section of the domain.
We find these using the First Derivative Test.

• Determine the critical points.
• Test intervals around these critical points to study the sign change of the derivative.
• If the derivative changes from positive to negative at a point, it's a relative maximum.
• If it changes from negative to positive, it's a relative minimum.

For \( f(x) = x^2 e^{-x} \), we applied the First Derivative Test and found:

• The derivative doesn't change sign at \( x = 0 \), indicating neither a relative maximum nor minimum.
• At \( x = 2 \), \( f'(x) \) changes from positive to negative, suggesting a relative maximum there.

This analysis through relative extrema helps us understand where the function peaks in its performance over specific intervals.