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Chapter 0: Problem 20

Find functions \(f\) and \(g\) such that \(h=g \circ f\) (Note: The answer is not unique.) \(h(x)=\sqrt{2 x+1}+\frac{1}{\sqrt{2 x+1}}\)

Short Answer

Expert verified
The short answer is: To find functions \(f\) and \(g\) such that \(h = g \circ f\), we can define \(f(x) = 2x + 1\) and \(g(u) = \sqrt{u} + \frac{1}{\sqrt{u}}\). With these functions, we have \(h(x) = g(f(x)) = \sqrt{2x+1} + \frac{1}{\sqrt{2x+1}}\).

Step by step solution

01

Break down the given function

The given function is: \[h(x) = \sqrt{2x+1} + \frac{1}{\sqrt{2x+1}}\] Let \(f(x) = 2x + 1\), then we can rewrite the given function as: \[h(x) = \sqrt{f(x)} + \frac{1}{\sqrt{f(x)}}\] Now, define a function \(g(u)\) such that \(u = f(x)\) and we have: \[g(u) = \sqrt{u} + \frac{1}{\sqrt{u}}\] So, now we have: \[h(x) = g(f(x))\] The functions we found are: \[f(x) = 2x + 1\] \[g(u) = \sqrt{u} + \frac{1}{\sqrt{u}}\] Therefore: \[h(x) = g(f(x)) = g(2x+1) = \sqrt{2x+1} + \frac{1}{\sqrt{2x+1}}\]

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