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Chapter 1: Problem 4

Check your answer by evaluating the appropriate function at your answer. Suppose \(g(x)=\frac{x-3}{x-4}\). Evaluate \(g^{-1}\) (2).

Short Answer

Expert verified
\(g^{-1}(2) = -5\)

Step by step solution

01

Find the inverse function

Given the function \(g(x) = \frac{x-3}{x-4}\), we need to find the inverse of g(x), which is denoted as \(g^{-1}(x)\). To find the inverse function, we can follow these steps: 1. Replace g(x) with y, which gives us the equation \(y = \frac{x-3}{x-4}\). 2. Swap x and y in the equation. 3. Solve for y.
02

Swap x and y in the equation

The equation \(y = \frac{x-3}{x-4}\) now becomes \(x = \frac{y-3}{y-4}\).
03

Solve for y

We need to solve the equation \(x = \frac{y-3}{y-4}\) for y. Let's follow these steps: 1. Multiply both sides by (y-4) to eliminate the denominator in the equation: x(y-4) = y-3 2. Distribute x: xy - 4x = y - 3 3. Bring y terms to one side and other terms to the opposite side: y - xy = 4x - 3 4. Factor y: y(1-x) = 4x - 3 5. Divide both sides by (1-x) to solve for y: \[y= \frac{4x-3}{1-x}\] Now, we have found the inverse function, \(g^{-1}(x) = \frac{4x-3}{1-x}\).
04

Evaluate g^(-1)(2)

We are asked to evaluate \(g^{-1}(2)\), so plug in x=2 into the inverse function: \(g^{-1}(2) = \frac{4(2)-3}{1-2} = \frac{5}{-1} = -5\) Our final answer is \[g^{-1}(2) = -5\].

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Key Concepts

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

Function Composition
Function composition involves combining two functions into a single operation. In this context, it helps in verifying inverse functions. A function \( f(x) \) and its inverse \( f^{-1}(x) \) should satisfy the condition of function composition: when \( f \) is applied to \( f^{-1} \), or vice versa, they produce the identity function.
  • If \( g(x) \) is a function and \( g^{-1}(x) \) its inverse, then \( g(g^{-1}(x)) = x \) for all \( x \) in the domain of \( g^{-1} \).
  • Similarly, \( g^{-1}(g(x)) = x \) for all \( x \) in the domain of \( g \).
In our case, if we apply \( g \) to its inverse, we should retrieve the original input. It's the reverse process of the original function. This property helps us to ensure the correctness of our inverse function. For instance, composing the functions for a test with \( g(g^{-1}(2)) \) should yield 2, the original input.
Rational Functions
Rational functions are expressions where both the numerator and the denominator are polynomials. They can often have different behaviors based on their inputs due to their algebraic form. Understanding rational functions makes it easier to handle them when exploring inverse functions.
  • The function \( g(x) = \frac{x-3}{x-4} \) has a numerator \( x-3 \) and a denominator \( x-4 \). This structure can introduce points of discontinuity, particularly where the denominator equals zero.
  • Inverting such functions involves careful variable interchange and algebraic manipulation to avoid undefined points.
It鈥檚 important to note, solving for an inverse involves ensuring you account for values that could cause the original function's denominator to be zero. These discontinuities are crucial when evaluating the function or its domain.
Solving Equations
Solving equations is about finding the values that satisfy an equation. This skill is central when finding inverse functions. You perform algebraic manipulations until you isolate the variable of interest.
  • With the equation \( x = \frac{y-3}{y-4} \), our goal is to solve for \( y \).
  • To clear the fraction, multiply both sides by \( y-4 \), resulting in \( x(y-4) = y-3 \).
  • Distribute and re-arrange to collect like terms, and factor as needed. This will help isolate \( y \).
  • Finally, divide through by the coefficient of \( y \) to solve completely, yielding the inverse function.
By diligently applying these techniques, you can derive \( y = \frac{4x-3}{1-x} \) as the inverse in a straightforward, systematic manner. Understanding these steps empowers you to deal with more complex algebraic structures.

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Most popular questions from this chapter

True or false: The product of an even function and an odd function (with the same domain) is an odd function. Explain your answer.

Find functions \(f\) and \(g\), each simpler than the given function \(h\), such that \(h=f \circ g\). \( h(x)=\frac{3}{2+x^{2}}\)

Suppose \(f\) and \(g\) are functions, each with domain of four numbers, with \(f\) and \(g\) defined by the tables below: $$\begin{array}{c|c}x & f(x) \\\\\hline 1 & 4 \\\2 & 5 \\\3 & 2 \\\4 & 3\end{array}$$ $$\begin{array}{c|c}x & g(x) \\\\\hline 2 & 3 \\\3 & 2 \\\4 & 4 \\\5 & 1\end{array}$$ Sketch the graph of \(g^{-1}\).

Find a number \(c\) such that \(f \circ g=g \circ f,\) where \(f(x)=5 x-2\) and \(g(x)=c x-3\) .

Suppose you are exchanging cur. rency in the London airport. The currency exchange service there only makes transactions in which one of the two currencies is British pounds, but you want to exchange dollars for Euros. Thus you first need to exchange dollars for British pounds, then exchange British pounds for Euros. At the time you want to make the exchange, the function \(f\) for exchanging dollars for British pounds is given by the formula \(f(d)=0.66 d-1\) and the function \(g\) for exchanging British pounds for Euros is given by the formula \(g(p)=1.23 p-2\) The subtraction of 1 or 2 in the number of British pounds or Euros that you receive is the fee charged by the currency exchange service for each transaction. Which process gives you more Euros: exchanging $$\$ 100$$ for Euros twice or exchanging $$\$ 200$$ for Euros once?
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