Question

In Exercises \(13-24,\) find \(f^{-1}\) and verify that $$\left(f \circ f^{-1}\right)(x)=\left(f^{-1} \circ f\right)(x)=x$$ \(f(x)=x^{2}+1, \quad x \geq 0\)

Short answer

The inverse of the function \(f(x) = x^{2} + 1\), \(x \geq 0\) is \(f^{-1}(x) = \sqrt{x - 1}\), and the properties \((f \circ f^{-1})(x) = x\) and \((f^{-1} \circ f)(x) = x\) are confirmed.

Step-by-step solution

Exchange \(x\) and \(y\)

To find the inverse of the function, we replace \(f(x)\) with \(y\). So, we have \(y = x^{2} + 1\). For an inverse, we exchange \(x\) and \(y\), so our equation becomes \(x = y^{2} + 1\). Next, solve for \(y\).

Solve for \(y\)

Start by subtracting 1 from both sides to isolate \(y^{2}\). This gives us \(x - 1 = y^{2}\). Since \(y = f^{-1}(x)\), and given the square root character, we only account for the positive root because \(x \geq 0\). Therefore, the inverse function is \(f^{-1}(x) = \sqrt{x - 1}\).

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

To confirm the first property of inverse functions given, we replace \(f^{-1}(x)\) in function \(f\) by \(f(f^{-1}(x)) = (\sqrt{x - 1})^{2} + 1 = x - 1 + 1 = x\). Thus, \(f \circ f^{-1}(x) = x\) is proven.

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

To validate the second property of inverse functions, we replace \(f(x)\) in function \(f^{-1}\) by \(f^{-1}(f(x)) = \sqrt{(x^{2}+1) - 1}= \sqrt{x^{2}} = x\). Therefore, \(f^{-1} \circ f(x) = x\) is accepted.