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Chapter 2: Problem 48

Difference Quotient Find \(f(a), f(a+h),\) and the difference quotient \(\frac{f(a+h)-f(a)}{h},\) where \(h \neq 0.\) $$f(x)=\frac{2 x}{x-1}$$

Short Answer

Expert verified
f(a) = \frac{2a}{a-1}, f(a+h) = \frac{2(a+h)}{a+h-1}, Difference Quotient: -\frac{2}{(a+h-1)(a-1)}

Step by step solution

01

Calculate f(a)

To find \(f(a)\), we substitute \(x\) with \(a\) in the function \(f(x) = \frac{2x}{x - 1}\). This gives us:\[f(a) = \frac{2a}{a - 1}\].
02

Calculate f(a + h)

Next, we substitute \(x\) with \(a + h\) into the function \(f(x) = \frac{2x}{x - 1}\). This gives us:\[f(a+h) = \frac{2(a+h)}{(a+h) - 1} = \frac{2(a+h)}{a+h-1}\].
03

Compute the Difference Quotient

We now have both \(f(a)\) and \(f(a+h)\). Substitute these into the difference quotient formula:\[\frac{f(a+h) - f(a)}{h}\].Substitute \(f(a+h)\) and \(f(a)\):\[\frac{\frac{2(a+h)}{a+h-1} - \frac{2a}{a-1}}{h}\].
04

Simplify the Difference Quotient

First, find a common denominator for the fractions inside the difference quotient. The common denominator is \((a+h-1)(a-1)\). Rewrite each fraction:\[\frac{2(a+h)(a-1) - 2a(a+h-1)}{(a+h-1)(a-1)h}\].Expand and simplify the numerator:- Expand: \[2(a+h)(a-1) - 2a(a+h-1) = 2(a^2 + ah - a - h) - 2(a^2 + ah - a)\].- Simplify: \[= 2a^2 + 2ah - 2a - 2h - (2a^2 + 2ah - 2a)\].- Cancel terms to: \[= -2h\].So the difference quotient is:\[\frac{-2h}{(a+h-1)(a-1)h}\].After cancelling \(h\) in the numerator and denominator,\[-\frac{2}{(a+h-1)(a-1)}\].

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Key Concepts

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

Function Evaluation
Function evaluation is the process of finding the value of a function for a specific input. In this exercise, we worked with the function \( f(x) = \frac{2x}{x-1} \). To evaluate this function for a particular value, say \( x = a \), we substitute \( a \) for \( x \) and simplify.
This gives us:
  • \( f(a) = \frac{2a}{a-1} \)
To evaluate \( f \) at another point, \( x = a + h \), we again substitute and simplify:
  • \( f(a+h) = \frac{2(a+h)}{a+h-1} \)
Why is this important? Understanding function evaluation is essential for tackling more complex problems, such as calculating the difference quotient. By finding \( f(a) \) and \( f(a+h) \), we prepare to derive the difference quotient, which shows us how the function changes at the point \( a \) as \( h \) approaches zero.
Rational Functions
A rational function is a type of function expressed as a ratio of two polynomials. In this exercise, our function \( f(x) = \frac{2x}{x-1} \) is a rational function because both the numerator and the denominator are polynomial expressions.
Rational functions have specific characteristics:
  • They can have vertical asymptotes, which occur where the denominator equals zero. For \( f(x) \), the vertical asymptote is at \( x = 1 \) because the denominator becomes zero.
  • They may have horizontal asymptotes determined by the degrees of the polynomials in the numerator and denominator.
  • They often display behavior that is not simple or linear, requiring careful evaluation.
Rational functions can represent various real-world phenomena and are important in calculus for understanding limits and continuity.
Precalculus Mathematics
Precalculus mathematics serves as the foundational skills needed to excel in calculus. The topics learned in precalculus, such as evaluating functions and understanding different types of functions like rational ones, are crucial.
The exercise in this context focuses on the concept of the difference quotient鈥攁 key idea in calculus鈥攚hich is used to find the derivative of a function at a point.
A difference quotient is calculated by:
  • Finding \( f(a) \) and \( f(a+h) \)
  • Substituting these into the formula \( \frac{f(a+h) - f(a)}{h} \)
  • Simplifying the expression to reveal how the function's value changes as \( h \) approaches zero
This step helps students transition from understanding static function behavior to analyzing dynamic changes, laying the groundwork for studying calculus.

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Most popular questions from this chapter

Suppose that $$h=f \circ g$$ If \(g\) is an even function, is \(h\) necessarily even? If \(g\) is odd, is \(h\) odd? What if \(g\) is odd and \(f\) is odd? What if \(g\) is odd and \(f\) is even?

DISCUSS: Finding an Inverse "in Your Head" In the margin notes in this section we pointed out that the inverse of a function can be found by simply reversing the operations that make up the function. For instance, in Example 7 we saw that the inverse of $$f(x)=3 x-2 \quad \text { is } \quad f^{-1}(x)=\frac{x+2}{3}$$ because the "reverse" of "Multiply by 3 and subtract 2" is "Add 2 and divide by 3 ". Use the same procedure to find the inverse of the following functions. (a) \(f(x)=\frac{2 x+1}{5}\) (b) \(f(x)=3-\frac{1}{x}\) (c) \(f(x)=\sqrt{x^{3}+2}\) (d) \(f(x)=(2 x-5)^{3}\) Now consider another function: $$f(x)=x^{3}+2 x+6$$ Is it possible to use the same sort of simple reversal of operations to find the inverse of this function? If so, do it. If not, explain what is different about this function that makes this task difficult.

A man is running around a circular track that is 200 m in circumference. An observer uses a stopwatch to record the runner's time at the end of each lap, obtaining the data in the following table. (a) What was the man's average speed (rate) between 68 s and 152 s? (b) What was the man's average speed between 263 s and 412 s? (c) Calculate the man's speed for each lap. Is he slowing down, speeding up, or neither? $$\begin{array}{|c|c|} \hline \text { Time (s) } & \text { Distance (m) } \\ \hline 32 & 200 \\ 68 & 400 \\ 108 & 600 \\ 152 & 800 \\ 203 & 1000 \\ 263 & 1200 \\ 335 & 1400 \\ 412 & 1600 \\ \hline \end{array}$$

The table gives the population in a small coastal community for the period 1997-2006. Figures shown are for January 1 in each year. (a) What was the average rate of change of population between 1998 and 2001 ? (b) What was the average rate of change of population between 2002 and 2004 ? (c) For what period of time was the population increasing? (d) For what period of time was the population decreasing? $$\begin{array}{|c|c|} \hline \text { Year } & \text { Population } \\ \hline 1997 & 624 \\ 1998 & 856 \\ 1999 & 1,336 \\ 2000 & 1,578 \\ 2001 & 1,591 \\ 2002 & 1,483 \\ 2003 & 994 \\ 2004 & 826 \\ 2005 & 801 \\ 2006 & 745 \\ \hline \end{array}$$

Find a function whose graph is the given curve. The line segment joining the points \((-2,1)\) and \((4,-6)\)
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