Question

In Exercises \(13-24,\) find \(f^{-1}\) and verify that $$\left(f \circ f^{-1}\right)(x)=\left(f^{-1} \circ f\right)(x)=x$$ \(f(x)=x^{2}+2 x+1, \quad x \geq-1\)

Short answer

The inverse function is \(f^{-1}(x) = \sqrt{x - 1} - 2\), and the compositions \(f \circ f^{-1}(x)\) and \(f^{-1} \circ f(x)\) both equal to \(x\).

Step-by-step solution

Find the Inverse Function

To find the inverse function, we can start by replacing \(f(x)\) with \(y\). So, we get \(y = x^2 + 2x + 1\). The aim is to solve for \(x\). We can rearrange the equation to \(x = (y-1)^2 - 2\). From this, the inverse function can be written as \(f^{-1}(x) = \sqrt{x - 1} - 2\).

Verify \(f \circ f^{-1}(x) = x\)

We need to substitute \(f^{-1}(x)\) into the equation for \(f(x)\), or replace \(x\) in \(f(x)\) with \(f^{-1}(x)\). Doing so gives us \(f(f^{-1}(x)) = (\sqrt{x - 1} - 2)^2 + 2(\sqrt{x - 1} - 2) + 1\). Simplifying this equation will lead to \(x\).

Verify \(f^{-1} \circ f(x) = x\)

Similarly, we need to substitute \(f(x)\) into the equation for \(f^{-1}(x)\), or replace \(x\) in \(f^{-1}(x)\) with \(f(x)\). Doing so gives us \(f^{-1}(f(x)) = \sqrt{(x^2 + 2x + 1) - 1} - 2\). Simplifying this equation will also lead to \(x\).