Problem 39 - College Algebra and Calculus: An Applied Approach

Question

In Exercises, find the absolute extrema of the function on the interval \([0, \infty)\). $$ f(x)=\frac{2 x}{x^{2}+4} $$

Step-by-Step Solution

Answer

To find the absolute extrema of a function, first derive the function, then find its critical points by setting the derivative equal to zero. Finally, evaluate the original function at these points as well as at the limit points of the given interval. The smallest and largest of these results are the absolute minimum and maximum of the function, respectively.

Step 1: Derive the Function

Take the derivative of the function \( f(x) \). Remember that we are dealing with a quotient, so we need to use the quotient rule which states that \((\frac{u}{v})' = \frac{u'v - uv'}{v^2}\). Apply this rule to \( f(x) = \frac{2x}{x^{2}+4} \) to get \( f'(x) \).

Step 2: Find the Critical Points

Set \( f'(x) = 0 \) and solve for 'x'. This will provide the critical points of the function where it might attain its maximum or minimum.

Step 3: Evaluate at the Critical and Boundary Points

Substitute each of the critical points and boundary points (in this case 0 and \( \infty \)) into the original function. The greatest and smallest of these values will be the absolute maximum and minimum of the function on the interval.

Key Concepts

Derivative

A derivative measures how a function changes as its input changes. It's like finding the slope of a curve at any given point. In our problem, we work with a quotient, which is a division of two functions. The function given, \( f(x) = \frac{2x}{x^2 + 4} \), is in quotient form, so we apply the quotient rule to derive it.

The **quotient rule** helps us differentiate functions like \( \frac{u}{v} \). The formula is:

• \( f'(x) = \frac{u'v - uv'}{v^2} \)

Plugging in the pieces for our function:

• \( u = 2x \), hence \( u' = 2 \)
• \( v = x^2 + 4 \), hence \( v' = 2x \)

Thus, the derivative \( f'(x) \) is calculated as

• \( f'(x) = \frac{2(x^2 + 4) - 2x(2x)}{(x^2 + 4)^2} \)

This derivative tells us the rate of change of the function \( f(x) \) with respect to \( x \). A crucial step towards finding critical points.

Critical Points

Critical points occur where the derivative of a function is zero or undefined. They are important because they might be where the function reaches a local maximum or minimum. In our equation, **setting the derivative equal to zero** helps us find these points.

To find the critical points, we solve:

• \( f'(x) = \frac{2(x^2 + 4) - 2x(2x)}{(x^2 + 4)^2} = 0 \)
• Since a fraction is zero only when its numerator is zero, consider \( 2(x^2 + 4) - 2x(2x) = 0 \)
• Simplify the equation: \( 2x^2 + 8 - 4x^2 = 0 \) leads to \( -2x^2 + 8 = 0 \)
• Solve for \( x \): \( x^2 = 4 \) gives us \( x = \pm 2 \)

Since we're only considering \([0, \infty)\), we focus on \( x = 2 \). This is a critical point where the function changes its behavior, possibly reaching an extremum.

Quotient Rule

The quotient rule is vital when dealing with derivatives of functions expressed as one function divided by another, like \( \frac{u}{v} \). It ensures accuracy when determining derivatives from complex rational functions.

The **quotient rule** states:

• \( (\frac{u}{v})' = \frac{u'v - uv'}{v^2} \)

This rule prioritizes the order and operations necessary for proper differentiation:

• First, differentiate the numerator (\( u \)) and the denominator (\( v \)) separately.
• Multiply the derivative of \( u \) by \( v \), then subtract the product of \( u \) and the derivative of \( v \).
• Lastly, divide the entire expression by \( v^2 \), the square of the denominator.

In our problem, this method converts the given function into a workable form to discover its behavior and to identify critical points. It effectively helps break down what might seem complex into simple calculable steps, aiding in the calculus of real-world problems.