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Chapter 4: Problem 69

Use a CAS to find the numbers \(x\) at which (a) \(f^{\prime}(x)=0, \quad\) (b) \(f^{\prime}(x) \cdot 0, \quad\) (c) \(f^{\prime}(x) \cdot 0.\) $$f(x)=\frac{x^{2}-1}{x^{2}+1}$$

Short Answer

Expert verified
The value of \(x\) that makes \(f'(x) = 0\) is \(x = 0\).

Step by step solution

01

Function Differentiation

Derive the function \(f(x)\) to find \(f'(x)\), using the quotient rule. The quotient rule is given as: \(h'(x) = \frac{g(x)f'(x) - g'(x)f(x)}{g^2(x)}\), where in this case, \(f(x) = x^2 - 1\) and \(g(x) = x^2 + 1\). Calculating \(f'(x)\), we get \(f'(x) = \frac{2x(x^2+1) - 2x(x^2-1)}{(x^2+1)^2}\).
02

Simplifying the Derivative

After cancelling out similar terms in the numerator, the function simplifies to: \(f'(x) = \frac{4x}{(x^2+1)^2}\)
03

Solving \(f'(x) = 0\)

Equating this derivative to zero i.e. \(f'(x) = 0\) yields: \(\frac{4x}{(x^2+1)^2} = 0\). The fraction is only zero when the numerator is zero. Therefore, the solution is when \(x = 0\).

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Critical Points
Critical points in calculus are locations on a graph where a function's derivative is zero or undefined. These points are critical because they signal potential peaks (maximums), troughs (minimums), or inflection points where the curve changes concavity.

How to find critical points:
  • First, find the derivative of the function. This derivative tells us how the function is changing at any given point.
  • Next, set this derivative equal to zero and solve for the variable. Any point where the derivative equals zero is a potential critical point.
  • Check if derivative is undefined for any x-values as these might also be critical points.
This process helps to determine where the function is changing behavior, as seen in the solution where solving for when the derivative of \( f(x) = \frac{x^2 - 1}{x^2 + 1} \) equals zero tells us \( x = 0 \) is a critical point.
Derivative
A derivative represents how a function changes as its input changes, essentially showing the rate of change or slope at any point on the graph. The derivative is a fundamental tool in calculus, serving as the foundation for understanding rates and optimizing functions.

Why derivatives are crucial:
  • Derivatives give us essential information about the behavior of functions, helping us locate critical points and understand increasing or decreasing trends.
  • They are used to find slopes of curves, letting us determine tangent lines at specific points.
In the original exercise, finding the derivative \( f'(x) = \frac{4x}{(x^2+1)^2} \) involved applying the quotient rule, showcasing how derivatives are calculated when a function is expressed as a ratio of two other functions.
Quotient Rule
When functions are expressed as a ratio, the quotient rule becomes indispensable in finding derivatives. The quotient rule helps differentiate functions given in the form \( \frac{f(x)}{g(x)} \). It's defined as:

\[ h'(x) = \frac{g(x)f'(x) - g'(x)f(x)}{g^2(x)} \]where:
  • \( f(x) \) is the function in the numerator
  • \( g(x) \) is the function in the denominator
Steps to applying the quotient rule:
  • Differentiate both the numerator and the denominator separately.
  • Apply the formula to calculate the derivative of the entire function.
  • Simplify the resulting expression to make calculations easier.
In our exercise, the quotient rule was applied to derive \( f'(x) \) for \( f(x) = \frac{x^2 - 1}{x^2 + 1} \), leading to a simplified derivative \( f'(x) = \frac{4x}{(x^2+1)^2} \). This illustrates how the quotient rule helps tackle complex rational functions more efficiently.

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Most popular questions from this chapter

The cost of erecting a small office building is 1,000,000 dollar for the first story, 1,100,000 dollar for the second, 1,200,000 dollar for the third, and so on. Other expenses (lot, basement, etc.) are \(5.5,000,000 .\) Assume that the annual rent is 200,000 dollar per story. How many stories will provide the greatest return on investment?

A revolving searchlight \(\frac{1}{2}\) mile from a straight shoreline makes I revolution per minute. How fast is the light moving along the shore as it passes over a shore point 1 mile from the shore point nearest to the scarchlight?

An object moves along a coordinate line, its position at each time \(t \geq 0\) being given by \(x(t)\). Find the times \(t\) at which the object changes direction. $$x(t)=\left(t^{2}-8 t+15\right)^{3}$$.

(Oblique asymplotes) Let \(r(x)=p(x) / q(x)\) be a rational function. If (degree of \(p)=(\text { degree of } q)+1,\) then \(r\) can be Written in the form \(r(x)=a x+b+\frac{Q(x)}{q(x)}\) with \((\text { degree } Q)<(\text { degree } q)\). Show that \([r(x)-(a x+b)] \rightarrow 0\) both as \(x \rightarrow \infty\) and as \(x \rightarrow-\infty .\) Thus the graph of \(f\) "approaches the line \(y=$$a x+b\)" both as \(x \rightarrow \infty\) and as \(x \rightarrow-\infty\). The line \(y=\) \(a x+b\) is called an oblique asymptote.

Sketch the graph of the function showing all vertical and oblique asymptotes. $$f(x)=\frac{x^{2}-4}{x}$$
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