Problem 52 - Thomas Calculus

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. $$ g(x)=\frac{x^{2}}{4-x^{2}}, \quad-2< x \leq 1 $$

Step-by-Step Solution

Answer

The local minimum is at \( x = 0 \), absolute minimum too. The endpoint at \( x = 1 \) is a maximum in the defined domain.

Step 1: Find the Derivative

To find the local extreme values, we start by finding the derivative of the function. The function is \( g(x) = \frac{x^2}{4-x^2} \). To find the derivative, use the quotient rule:\[ g'(x) = \frac{(4-x^2) \cdot 2x - x^2 \cdot (-2x)}{(4-x^2)^2} = \frac{8x-4x^3}{(4-x^2)^2}. \]

Step 2: Set the Derivative to Zero

Local extrema occur where the derivative is zero or undefined. Setting the derivative to zero:\[ \frac{8x - 4x^3}{(4-x^2)^2} = 0 \Rightarrow 8x - 4x^3 = 0. \]Factor out common terms:\[ 4x(2 - x^2) = 0. \]This gives solutions: \( x = 0 \) and \( x = \pm\sqrt{2} \). Since the interval is \(-2 < x \leq 1\), only \( x = 0 \) is valid.

Step 3: Evaluate Critical Points and End Points

Evaluate the function at the critical point within the domain and the endpoints of the domain:\[ g(0) = \frac{0^2}{4-0^2} = 0, \]\[ g(1) = \frac{1^2}{4-1^2} = \frac{1}{3}. \]Check the function behavior as \( x \to -2^+ \); although it's approaching an asymptote, it's not explicitly defined, so we check only within open values slightly greater than -2.

Step 4: Determine Local and Absolute Extrema

The function has a local minimum at \( x=0 \) with a value of \( g(0) = 0 \). Since this is the smallest value observed and it occurs within the domain, it is also an absolute minimum. At the endpoint \( x = 1 \), the value is \( g(1) = \frac{1}{3} \). The function tends to a large positive number as \( x \to -2^+ \); however, this is not within the considered discrete domain for a starting point.

Step 5: Verify with a Graph

Plot the function \( g(x) \) using a graphing tool within the domain \(-2 < x \leq 1\). Verify that the curve demonstrates a local minimum at \( x = 0 \) and increases towards \( x = 1 \). Check the behavior as it approaches the asymptote from the open left endpoint.

Key Concepts

Understanding Derivatives in Finding Extrema

When solving problems involving local extrema, the derivative of a function is your first stop. A derivative, essentially, measures how a function is changing at any given point and is denoted as \( g'(x) \) for our function \( g(x) = \frac{x^2}{4-x^2} \). Finding the derivative involves knowing rules like the power rule, product rule, and particularly here, the quotient rule. Once you have the derivative, set it to zero to identify critical points.Think of these points as the locations where the graph momentarily "turns around," potentially indicating a peak or a trough. However, don't forget to check points where the derivative is undefined, as these too can tell interesting stories about the graph's behavior. To fully understand change within a specific interval, as required by the problem here, you then verify the values of these points against the endpoints. This process helps in identifying whether these points represent local minima or maxima in the defined domain.

The Quotient Rule Explained

The quotient rule is your go-to tool when tackling derivatives of ratios, which is the structure of our function \( g(x) = \frac{x^2}{4-x^2} \). It states that if you have a function \( h(x) = \frac{u(x)}{v(x)} \), its derivative is defined as \( h'(x) = \frac{v(x)u'(x) - u(x)v'(x)}{(v(x))^2} \). Applied to our function:

• \( u(x) = x^2 \) and \( v(x) = 4 - x^2 \)
• \( u'(x) = 2x \) and \( v'(x) = -2x \)

Insert these into the formula, simplify, and you get the derivative: \( g'(x) = \frac{8x - 4x^3}{(4-x^2)^2} \).At this point, the critical task is analyzing values where this derivative equals zero, giving us potential extrema. It's insightful to practice using the quotient rule, as it frequently arises in calculus for more intricate functions.

Determining Absolute Extremum

An absolute extremum is the point where the function reaches its highest or lowest value over the entire domain. Once you've found where the derivative equals zero (critical points), you need to compare these with the values at the domain's endpoints to identify if they hold the smallest or largest value.For the function \( g(x) \) on the interval \(-2 < x \leq 1\), after evaluating the derivative, you'll find that \( x = 0 \) gives a value of \( g(0) = 0 \), which turns out to be not just a local minimum but the absolute minimum as well. The key is to confirm this by checking all critical points within the interval and endpoints, ensuring there's no lower value elsewhere. If the value at the endpoint \( x = 1 \) were greater, it would not qualify as the absolute minimum.

Using a Graphing Calculator for Verification

To solidify your findings and ensure there are no errors, using a graphing calculator can be incredibly beneficial. It will visually display the function \( g(x) = \frac{x^2}{4-x^2} \) over the interval \(-2 < x \leq 1\). By setting your calculator to this domain, you can see if the behavior matches your critical point analysis.A graphing tool will also help illustrate how the function approaches the asymptote near \( x = -2 \) and verify whether your pinpointed extremum at \( x=0 \) accurately represents the lowest point of the graphical curve. Besides confirming the calculations, graphing gives you a robust understanding of how functions behave, especially useful when preparing for similar calculus problems in the future. These visual aids enhance your analytical approach to solving extrema problems.