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Chapter 3: Problem 22

(a) find the intervals on which \(f\) is increasing or decreasing, and (b) find the relative maxima and relative minima of \(\vec{f}\). $$ f(x)=\frac{x}{x-1} $$

Short Answer

Expert verified
The function \(f(x) = \frac{x}{x-1}\) is decreasing on the intervals \((-\infty, 1)\) and \((1, \infty)\). There are no relative maxima or minima in the function.

Step by step solution

01

Find the derivative of the function

To differentiate f(x) = \(\frac{x}{x - 1}\), we can use the quotient rule. If u(x) = x and v(x) = x - 1, the quotient rule states: $$ f'(x) = \frac{u'(x)v(x) - u(x)v'(x)}{v(x)^2} $$ Now, let's find the derivative of f(x).
02

Calculating the derivative using the quotient rule

Calculate the derivatives of u(x) and v(x) and plug them into the formula to find f'(x): u'(x) = \(\frac{d}{dx}\) (x) = 1 v'(x) = \(\frac{d}{dx}\) (x - 1) = 1 Now, plug these derivatives and u(x), v(x) into the formula: $$ f'(x) = \frac{1(x - 1) - x(1)}{(x - 1)^2} = \frac{x - 1 - x}{(x - 1)^2} = \frac{-1}{(x - 1)^2} $$
03

Determine the intervals on which f is increasing or decreasing

With the derivative found, we can now determine where the function is increasing and decreasing. Since the denominator \((x - 1)^2\) is always positive, with the exception of x = 1 when it is undefined, f'(x) changes its sign based on the numerator (-1) only: f'(x) < 0 for all x 鈮� 1 In conclusion, f(x) is decreasing on the intervals \((-\infty, 1)\) and \((1, \infty)\), and the function is not increasing on any interval.
04

Identify the relative maxima and minima

Since f(x) is decreasing on all intervals for x 鈮� 1, there are no relative maxima or minima in this function. The function is continuous and strictly decreasing for all x values except 1.

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Key Concepts

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

Quotient Rule
The quotient rule is a fundamental tool in calculus for finding the derivative of a function that is the quotient of two other functions. This rule comes in handy when dealing with a function expressed as the division of one function by another, like \( f(x) = \frac{u(x)}{v(x)} \). In such cases, the derivative \( f'(x) \) can be calculated using the formula:\[f'(x) = \frac{u'(x)v(x) - u(x)v'(x)}{v(x)^2}\]
  • \( u(x) \) and \( v(x) \) are the numerator and denominator functions, respectively.
  • \( u'(x) \) and \( v'(x) \) are the derivatives of these functions.
  • The denominator in the formula, \( v(x)^2 \), ensures the function remains valid except where \( v(x) = 0 \), so discontinuities must be considered.
Using this rule, we can reliably find the derivative of complex fractions in calculus problems.
Derivative
Derivatives provide a way to understand the rate of change of functions. In a practical sense, derivatives help identify how a function's output value changes concerning its input. When we derive \( f(x) = \frac{x}{x-1} \), the quotient rule gives us:\[f'(x) = \frac{-1}{(x-1)^2}\]This tells us several important things:
  • The derivative is negative, indicating that the function is decreasing.
  • It is undefined when \( x = 1 \), where the original function is also undefined.
By examining the sign of \( f'(x) \) across different intervals, you can determine the behavior of the function across its domain.
Increasing and Decreasing Intervals
Determining where a function is increasing or decreasing is essential for understanding its overall behavior. We achieve this by analyzing the sign of the derivative.
  • If \( f'(x) > 0 \), the function is increasing.
  • If \( f'(x) < 0 \), the function is decreasing.
For the function \( f(x) = \frac{x}{x-1} \), we've already calculated that \( f'(x) = \frac{-1}{(x-1)^2} \), which is negative for all \( x eq 1 \). Thus, the function is decreasing on both intervals \((-\infty, 1)\) and \((1, \infty)\). This means there are no intervals where the function is increasing.
Relative Maxima and Minima
Finding relative maxima and minima involves examining where a function changes its increasing or decreasing behavior. These points are crucial in optimization problems and understanding the function's graph shape. Generally:
  • Relative maxima occur where the function shifts from increasing to decreasing.
  • Relative minima occur where the function shifts from decreasing to increasing.
In our case, \( f(x) = \frac{x}{x-1} \) is strictly decreasing, indicating no change in behavior that would create relative extrema. This implies that there are no points of relative maxima or minima in the defined intervals. As a result, the function consistently trends downward, reaffirming its decreasing nature.

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Most popular questions from this chapter

Air Pollution The level of ozone, an invisible gas that irritates and impairs breathing, that is present in the atmosphere on a certain day in June in the city of Riverside is approximated by $$ S(t)=1.0974 t^{3}-0.0915 t^{4} \quad 0 \leq t \leq 11 $$ where \(S(t)\) is measured in Pollutant Standard Index (PSI) and \(t\) is measured in hours with \(t=0\) corresponding to 7 A.M. Plot the graph of \(S\), and interpret your results.

Use Newton's method to find the point of intersection of the graphs to four decimal places of accuracy by solving the equation \(f(x)-g(x)=0 .\) Use the initial estimate \(x_{0}\) for the \(x\) -coordinate. f(x)=\sin x, g(x)=\frac{1}{5} x, \quad x_{0}=2

The graph of \(f(x)=x-\sqrt{1-x^{2}}\) accompanies Exercise 4 . Explain why \(x_{0}=1\) cannot be used as an initial estimate for solving the equation \(f(x)=0\) using Newton's method. Can you explain this analytically? How about the initial estimate \(x_{0}=0 ?\)

Worldwide PC Shipments The number of worldwide PC shipments (in millions of units) from 2005 through 2009 , according to data from the International Data Corporation, are given in the following table. \begin{tabular}{|l|c|c|c|c|c|} \hline Year & 2005 & 2006 & 2007 & 2008 & 2009 \\ \hline PCs & \(207.1\) & \(226.2\) & \(252.9\) & \(283.3\) & \(302.4\) \\ \hline \end{tabular} By using the logistic curve-fitting capability of a graphing calculator, it can be verified that a regression model for this data is given by $$ f(t)=\frac{544.65}{1+1.65 e^{-0.1846 r}} $$ where \(t\) is measured in years and \(t=0\) corresponds to 2005 . a. Plot the scatter diagram and the graph of the function \(f\) using the viewing window \([0,4] \times[200,300]\). b. How fast were the worldwide PC shipments increasing in \(2006 ?\) In \(2008 ?\)

In Exercises \(55-58\), plot the graph of the function. $$ f(t)=\frac{\sqrt{t^{2}+1}}{t-1} $$
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