Problem 22 - Calculus Early Transcendentals: Pearson New International Edition

Question

Find, if possible, the (global) maximum and minimum values of the given function on the indicated interval. $$ f(x)=\frac{2 x}{x^{2}+4} \text { on }[0, \infty) $$

Step-by-Step Solution

Answer

The global maximum is \( \frac{1}{2} \) at \( x=2 \), and the minimum is \( 0 \) at \( x=0 \) and as \( x \rightarrow \infty \).

Step 1: Find the Derivative

To find the critical points of the function \( f(x) = \frac{2x}{x^2 + 4} \), we first need to find the derivative \( f'(x) \). We will use the quotient rule, which states if \( f(x) = \frac{g(x)}{h(x)} \), then \( f'(x) = \frac{g'(x)h(x) - g(x)h'(x)}{(h(x))^2} \). Here, \( g(x) = 2x \) and \( h(x) = x^2 + 4 \). Thus, \( g'(x) = 2 \) and \( h'(x) = 2x \). Substituting these into the quotient rule gives:\[f'(x) = \frac{2(x^2 + 4) - 2x(2x)}{(x^2 + 4)^2} = \frac{8 - 2x^2}{(x^2 + 4)^2}.\]

Step 2: Find Critical Points

Set the derivative \( f'(x) = 0 \) to find critical points. Solve for \( x \):\[\frac{8 - 2x^2}{(x^2 + 4)^2} = 0 \implies 8 - 2x^2 = 0.\]Rearrange and solve the equation:\[2x^2 = 8 \quad \Rightarrow \quad x^2 = 4 \quad \Rightarrow \quad x = 2\ ext{or} \ x = -2.\]Since the interval is \([0, \infty)\), we only consider \( x = 2 \).

Step 3: Evaluate the Function at Critical Points and Endpoints

Compute \( f(x) \) at the endpoints and critical point \( x = 2 \). The interval is \([0, \infty)\), so the endpoint is \( x = 0 \). Evaluate:\[f(0) = \frac{2 \times 0}{0^2 + 4} = 0, \quad f(2) = \frac{2 \times 2}{2^2 + 4} = \frac{4}{8} = \frac{1}{2}.\]Check limits as \( x \rightarrow \infty \):\[ f(x) = \frac{2x}{x^2 + 4} \approx \frac{2x}{x^2} = \frac{2}{x} \rightarrow 0.\]

Step 4: Determine Global Maximum and Minimum

The function values at critical and endpoint are \( f(0) = 0 \), \( f(2) = \frac{1}{2} \), and \( f(x) \rightarrow 0 \) as \( x \rightarrow \infty \). The maximum value is \( \frac{1}{2} \) at \( x = 2 \), and the minimum value is \( 0 \) occurring as \( x \rightarrow \infty \) and at \( x = 0 \).

Key Concepts

Global Maximum

In calculus, determining a global maximum of a function is all about finding the highest point over its entire interval. When we talk about global maximum in the context of calculus optimization, we aim to locate the maximum output that a function can achieve over a specified range.
This is different from a local maximum, which is simply the highest point within a small neighborhood.
In the problem we are considering, the function is examined over the interval \([0, \infty)\).

• A global maximum value is when the function has a peak that surpasses all other function values within the given interval.
• In our example, the global maximum is found by evaluating the function at critical points and endpoints, as these are key spots where extreme values commonly occur.
• For the function \( f(x) = \frac{2x}{x^2 + 4} \), we calculated the global maximum to be \( \frac{1}{2} \) at \( x = 2 \).

Finding a global maximum is crucial in fields requiring optimization, such as when a business seeks to maximize profits within constraints.

Global Minimum

Just as with finding a global maximum, we can determine a global minimum by examining the lowest point of a function over its complete interval. A global minimum is where the function attains its smallest value on the set interval.
For the function in our exercise \( f(x) = \frac{2x}{x^2 + 4} \), we need to evaluate the function at the interval's endpoints and critical points.

• A critical feature of global minimum calculations is the need to consider limiting behavior as shown when \(x \rightarrow \infty \) where the function approaches \(0\).
• Evaluating \(f(x)\) at the endpoints might tell us about minimums, specifically at \(x = 0\), this value is \(0\).
• Thus, in this problem, the global minimum value is \(0\), occurring both at \(x = 0\) and as \(x\) heads to infinity.

The importance of determining global minimum values cannot be understated, especially where minimizing costs or resources is needed.

Critical Points

Critical points are key in identifying function behaviors, particularly where a function might achieve local or global maxima or minima.
A critical point occurs at \(x\) where the derivative of a function equals zero or does not exist. Determining these can help us understand significant trends and changes in the function's slope.

• In our step-by-step problem, we found critical points by setting the derivative to zero,\( f'(x) = 0 \).
• Our computations gave \( x = 2\) as a critical point within the given interval \([0, \infty)\).
• These points are vital as they mark where the function may stop increasing or decreasing.

To ensure clarity, looking at the sign of the derivative around these points aids in predicting behavior, specifically whether the point is a maximum or a minimum.

Quotient Rule

When dealing with functions expressed as a ratio, the quotient rule is the go-to method for finding derivatives.
The quotient rule helps calculate the derivative of a function that can be written in the form \( \frac{g(x)}{h(x)} \), where both \(g(x)\) and \(h(x)\) are differentiable functions.
Let's dissect the rule:

• For a function \( f(x) = \frac{g(x)}{h(x)} \), the derivative \( f'(x) = \frac{g'(x)h(x) - g(x)h'(x)}{(h(x))^2} \).
• Applying it to our problem: \( g(x) = 2x \) and \( h(x) = x^2 + 4 \), we computed \( f'(x) = \frac{8 - 2x^2}{(x^2 + 4)^2} \).
• This derivative informs us about the function's rate of change, revealing critical points when set to zero.

One must remember that the quotient rule often requires neat algebraic manipulation to arrive at a simplified and actionable derivative for further analysis.