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Chapter 8: Problem 17

In Exercises \(16-22,\) show that the two functions are inverses of each other. $$ f(x)=\sqrt{x-1} \text { (where } x>1 \text { ) and } g(x)=x^{2}+1 \text { (where } \left.x>0\right) $$

Short Answer

Expert verified
f(g(x)) = x and g(f(x)) = x.

Step by step solution

01

Understand the problem

To show that two functions are inverses of each other, we need to demonstrate that their compositions result in the identity function. This means showing that \(f(g(x)) = x\) and \(g(f(x)) = x\).
02

Find \(f(g(x))\)

First, substitute \(g(x)\) into \(f(x)\). Recall \(g(x) = x^2 + 1\), so \(f(g(x)) = f(x^2 + 1)\). Since \(f(x) = \sqrt{x-1}\), we have: \ f(x^2 + 1) = \sqrt{(x^2 + 1) - 1} = \sqrt{x^2} = x\.
03

Find \(g(f(x))\)

Next, substitute \(f(x)\) into \(g(x)\). Recall \(f(x) = \sqrt{x - 1}\), so \(g(f(x)) = g(\sqrt{x - 1})\). Since \(g(x) = x^2 + 1\), we have: \ g(\sqrt{x - 1}) = (\sqrt{x - 1})^2 + 1 = x - 1 + 1 = x\.
04

Conclusion

Since both \(f(g(x)) = x\) and \(g(f(x)) = x\), we have shown that the functions \(f(x) = \sqrt{x-1}\) and \(g(x) = x^2 + 1\) are indeed inverses of each other.

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Key Concepts

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

Function Composition
Function composition refers to applying one function to the results of another function. For example, if we have two functions, \(f(x)\) and \(g(x)\), we find their composition by substituting the output of one function into the input of the other. This means that for the functions \(f(x) = \sqrt{x-1}\) and \(g(x) = x^2 + 1\), we perform:\[ f(g(x)) = f(x^2 + 1) \quad \text{and} \quad g(f(x)) = g(\sqrt{x-1}) \] When calculating \(f(g(x))\), we replace \(g(x)\) into \(f(x)\). Similarly, for \(g(f(x))\), we replace \(f(x)\) into \(g(x)\). This leads us to verify whether the composition results in the identity function.
Identity Function
An identity function is a function that, when applied to any input, returns the same input. Mathematically, the identity function can be written as \(I(x) = x\). To show that two functions are inverses, you must prove that the composition of these functions equals the identity function for both input scenarios: \(f(g(x)) = x\) and \(g(f(x)) = x\). This means substituting one function into another, simplifying, and confirming that the output matches the original input.
Square Root Function
The square root function, denoted as \(f(x) = \sqrt{x}\ \), takes a non-negative number and returns its principal (non-negative) square root. For instance, \(f(9) = \sqrt{9} = 3\). When dealing with inverse functions, it's important to remember the domains and ranges. For our example functions, \(f(x) = \sqrt{x-1}\), the domain is \(x > 1\), ensuring the value inside the square root is non-negative, and the range is \(y > 0\). This ensures the final outputs are meaningful and real.
Quadratic Function
A quadratic function is a polynomial function of degree 2, generally given by \(g(x) = ax^2 + bx + c\). In our case, we use \(g(x) = x^2 + 1\). The quadratic function is shaped as a parabola and has important properties, such as having one vertex and being symmetric about its axis. The domain is all real numbers, but in the context of inverse functions, we may need to restrict it to maintain one-to-one correspondence. For instance, here the range of \(g(x) = x^2 + 1\) is \(y \geq 1\), meaning we only deal with positive \(x\) values to find valid inverses effectively. For this reason, the inverse function must accurately map the outputs back to the correct domain inputs.

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Most popular questions from this chapter

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In Exercises \(16-22,\) show that the two functions are inverses of each other. $$ f(x)=2 x-1 \text { and } g(x)=\frac{x+1}{2} $$
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