Problem 20 - Calculus: Early Transcendentals

Question

Find the absolute extrema for \(f(x)=x e^{-x^{2} / 32}\) on [0,2] .

Step-by-Step Solution

Answer

The absolute extrema are maximum at \( x=2 \) with value \( 2e^{-1/8} \), and minimum at \( x=0 \) with value 0.

Step 1: Understand the Problem

We need to find both the maximum and minimum values of the function \( f(x) = x e^{-x^{2}/32} \) within the interval [0,2]. These are known as the absolute extrema.

Step 2: Find the Critical Points

Calculate the derivative of the function \( f(x) \). Use the product rule: if \( u = x \) and \( v = e^{-x^2/32} \), then \( f'(x) = u'v + uv' \). This gives \( f'(x) = 1 \cdot e^{-x^{2}/32} - x \cdot \frac{2x}{32} e^{-x^{2}/32} \). Simplify this to get \( f'(x) = e^{-x^{2}/32}(1 - \frac{x^2}{16}) \). Set \( f'(x) = 0 \) to find critical points: \( 1 - \frac{x^2}{16} = 0 \). Solving for \( x \) gives \( x^2 = 16 \) or \( x = 4 \) and \( x = -4 \). Only consider \( x = 4 \) since \( x = -4 \) is outside the domain [0,2].

Step 3: Evaluate the Endpoints

To find absolute extrema, evaluate \( f(x) \) at the endpoints of the interval. Calculate \( f(0) = 0 \cdot e^{-0^2/32} = 0 \) and \( f(2) = 2e^{-2^2/32} = 2e^{-1/8} \).

Step 4: Verify the Validity of Critical Points

Check if any critical points found in Step 2 fall within the interval [0, 2]. Since \( x = 4 \) is not within [0, 2], it is not considered.

Step 5: Determine the Absolute Extrema

Now, simply compare the function values at the critical points and endpoints to find the absolute maximum and minimum. Since critical points within [0,2] are not valid, only consider \( f(0) = 0 \) and \( f(2) = 2e^{-1/8} \). \( f(2) \) is larger than \( f(0) \), so the absolute maximum is at \( x = 2 \). The minimum must be \( f(0) = 0 \), as no other points are in the domain.

Key Concepts

Understanding Critical Points

Critical points are essential in identifying the extremum values of a function within a given interval. A critical point occurs where the derivative of the function equals zero or is undefined. These are points where the function may change direction, indicating potential local maxima or minima.
To find the critical points for a function, calculate its derivative and solve for zero. In our case, we find that \(f'(x)\) simplifies to \(e^{-x^{2}/32}(1 - \frac{x^2}{16})\).
Setting \(f'(x) = 0\) gives us the values of \(x\) that are potential critical points. For the function \(f(x) = xe^{-x^2/32}\), solving \(1 - \frac{x^2}{16} = 0\) yields \(x = 4\) and \(x = -4\). Since we are only interested in the interval [0, 2], these values are not applicable for absolute extrema.

Applying the Product Rule

The product rule is a vital tool in calculus for differentiating functions that are the product of two or more functions. It states that if a function \(y = uv\), where \(u\) and \(v\) are functions of \(x\), then the derivative \(y'\) is given by \(u'v + uv'\).
Let us apply this to our function \(f(x) = xe^{-x^2/32}\). Here, \(u = x\) and \(v = e^{-x^2/32}\). The derivatives are \(u' = 1\) and \(v' = -\frac{2x}{32} e^{-x^2/32}\).
Substituting these into the product rule formula, we obtain \(f'(x) = e^{-x^{2}/32} - x\frac{2x}{32}e^{-x^2/32}\), illustrating how product rule helps break down complex differentiation problems.

Derivative Calculation Insights

Calculating derivatives is central to finding critical points and understanding the behavior of a function. Differentiation gives us the rate of change of the function, hinting where it may reach extremum values.
For \(f(x) = xe^{-x^2/32}\), applying the derivative calculation yields \(f'(x) = e^{-x^{2}/32}(1 - \frac{x^2}{16})\). This result tells us how the function changes at different values of \(x\).

• The term \(e^{-x^2/32}\) reflects the exponential decrease or growth.
• \(1 - \frac{x^2}{16}\) indicates a balancing factor that adjusts the growth based on the value of \(x\).

Adjusting these elements allows us to pinpoint where the function increases, decreases, or reaches an extremum.

Evaluating Endpoints for Absolute Extrema

To find absolute extrema, it's crucial to evaluate the function at the interval's endpoints. Endpoints can be where the function achieves its highest or lowest values, even if no critical points are valid within the interval.
For \(f(x)=xe^{-x^2/32}\) in the interval [0, 2], it's necessary to calculate \(f(0)\) and \(f(2)\).

• At \(x = 0\), the function results in \(f(0) = 0\).
• At \(x = 2\), the function evaluates to \(f(2) = 2e^{-1/8}\).

.Compare these values to establish that the absolute maximum occurs at \(x = 2\) and the absolute minimum at \(x = 0\). This process demonstrates how endpoints often reveal extrema, especially when no suitable critical points exist within the closed interval.