Problem 56 - University Calculus: Early Transcendentals

Question

a. Identify the function's local extreme values in the given domain, and say where they occur. b. Which of the extreme values, if any, are absolute? c. Support your findings with a graphing calculator or computer grapher. $$g(x)=\frac{x^{2}}{4-x^{2}}, \quad-2

Step-by-Step Solution

Answer

The local minimum is at \(x = 0\); no absolute maximum by definition.

Step 1: Find the Derivative

To find the critical points where local extrema may occur, we first need to find the derivative of the function. Given \( g(x)=\frac{x^2}{4-x^2} \), use the quotient rule for derivatives: \[ g'(x) = \frac{d}{dx} \left( \frac{x^2}{4-x^2} \right) = \frac{(4-x^2)(2x) - (x^2)(-2x)}{(4-x^2)^2} = \frac{8x}{(4-x^2)^2}. \]

Step 2: Find Critical Points

Critical points occur where the derivative is zero or undefined. Set the derivative equal to zero: \[ 8x = 0. \] This implies that \( x = 0 \). We also check where the derivative might be undefined; since \( (4-x^2)^2 \) is in the denominator, check where it is zero, but this would occur at \( x = \pm 2 \), which are outside or at the boundary of the domain \( -2 < x \leq 1 \). Thus, only \( x = 0 \) is a critical point in the domain.

Step 3: Use Second Derivative Test

To classify the critical point at \( x = 0 \), use the second derivative test. First, find the second derivative: \[ g''(x) = \frac{d}{dx} \left( \frac{8x}{(4-x^2)^2} \right). \] Through computations, get: \[ g''(x) = \frac{8(4+x^2)}{(4-x^2)^3}. \] Substitute \( x = 0 \): \[ g''(0) = \frac{32}{4^3} = \frac{32}{64} = \frac{1}{2}, \] which is positive, indicating a local minimum at \( x = 0 \).

Step 4: Evaluate Function at Critical and Boundary Points

Evaluate the function at the critical point and endpoints of the domain to identify potential absolute extrema: - \( g(0) = \frac{0^2}{4-0^2} = 0 \)- \( g(1) = \frac{1^2}{4-1^2} = \frac{1}{3} \)The function is undefined at the lower bound \( x = -2 \), but asymptotically approaches positive values near \( x \to -2^+ \). Thus, potential extreme valuesare \( g(0) = 0 \) and \( g(1) = \frac{1}{3} \).

Step 5: Analyze the Graph

Use a graphing calculator or software to sketch \( g(x) = \frac{x^2}{4-x^2} \) on \( -2 < x \leq 1 \). Observe that the local minimum at \( x = 0 \) matches and there's an increasing tendency towards \( x = 1 \). The graph helps confirm that \( g(0) = 0 \) is a local and absolute minimum, and \( g(1) = \frac{1}{3} \) is a boundary value thought to potentially be an absolute maximum.

Key Concepts

Quotient Rule for Derivatives

In calculus, the quotient rule is essential when finding the derivative of a function represented as a fraction or a quotient of two differentiable functions. The formula is helpful when we need to differentiate a function like \( g(x) = \frac{f(x)}{h(x)} \). To apply the quotient rule, remember:

• The rule is given by: \( \frac{d}{dx}\left( \frac{f(x)}{h(x)} \right) = \frac{f'(x)h(x) - f(x)h'(x)}{[h(x)]^2} \).
• This means you take the derivative of the numerator \( f(x) \) and multiply it by the denominator \( h(x) \).
• Then, subtract the numerator \( f(x) \) multiplied by the derivative of the denominator \( h'(x) \).
• Finally, divide everything by the square of the denominator \( [h(x)]^2 \).

In our function \( g(x) = \frac{x^2}{4-x^2} \), applying the quotient rule allows us to find the derivative \( g'(x) = \frac{8x}{(4-x^2)^2} \), which is crucial for determining the critical points.

Critical Points

Identifying critical points is a vital step in analyzing a function's behavior, especially when looking for local extrema. A critical point occurs where the derivative of a function is zero or undefined. Here’s how we find them:

• First, find the derivative of the function. For \( g(x) = \frac{x^2}{4-x^2} \), we found \( g'(x) = \frac{8x}{(4-x^2)^2} \).
• Next, set the derivative equal to zero. Solving \( 8x = 0 \) gives us \( x = 0 \).
• Also, check where the derivative is undefined. Since the denominator \( (4-x^2)^2 \) could potentially be zero at \( x = \pm 2 \), these values are checked for undefined behavior, though here they lie outside or on the boundary of the given domain.

In the domain \( -2 < x \leq 1 \), the only critical point is \( x = 0 \), where the derivative is zero, suggesting potential local extremum behavior.

Second Derivative Test

The second derivative test is a handy tool for classifying critical points as local minima or maxima. Once you identify the critical points, follow these steps:

• Compute the second derivative of your function. For our function, it becomes \( g''(x) = \frac{8(4+x^2)}{(4-x^2)^3} \).
• Evaluate the second derivative at the critical point \( x = 0 \). Here, \( g''(0) = \frac{1}{2} \), a positive value.
• If the second derivative at the critical point is positive, the function has a local minimum at that point. If negative, it's a local maximum. A zero value suggests further investigation is necessary.

Thus, for \( x = 0 \), the second derivative gives us a positive value, indicating a local minimum.

Graphing Calculator

Graphing calculators are powerful tools that allow you to visualize functions and their properties comprehensively. They can be especially useful for validating analytical findings like local extrema. Here's how you can use one:

• Enter the function \( g(x) = \frac{x^2}{4-x^2} \) into the calculator.
• Set the graphing window to match the function's domain \( -2 < x \leq 1 \).
• Observe the behavior of the graph at the critical point and boundaries.

The graph should display a clear local minimum at \( x = 0 \) and show the function increasing as it approaches \( x = 1 \). This visual check confirms our analytical work regarding local extreme values.

Function Analysis

Function analysis involves thoroughly understanding how a function behaves over its domain. This includes:

• Determining critical points to locate potential extrema.
• Using the second derivative to classify these extrema as minima or maxima.
• Evaluating the function at critical and boundary points to identify absolute extrema.

For our function \( g(x) \):

• We located a critical point at \( x = 0 \), which is a local (and absolute) minimum since \( g(0) = 0 \).
• The function's behavior near boundaries \( x = -2 \) and \( x = 1 \) helped us identify a potential absolute maximum at \( g(1) = \frac{1}{3} \) as it approaches \( x = 1 \) from the left.

Analyzing the function in this manner provides a clear picture of its behavior and extremum values within the specified domain.