Question

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

Short answer

The inverse function of \(f(x)=x^{2}, x \leq 0\) is \(f^{-1}(x) = -\sqrt{x}\), and both \(f(f^{-1}(x)) = x\) and \(f^{-1}(f(x)) = x\) are true, as required.

Step-by-step solution

Find the Inverse

To find the inverse of a function, you need to swap \(x\) and \(y\), and then solve for \(y\). For the function \(f(x) = x^2\), first let \(y = f(x) = x^2\). Now swap \(x\) and \(y\), getting \(x = y^2\). Solving for \(y\) (square root of \(x\)), you get \(y = f^{-1}(x) = \pm \sqrt{x}\). But since \(x \leq 0\), the \(+\) root does not apply. So, the inverse function of \(x^2\) is \(f^{-1}(x) = -\sqrt{x}\).

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

Take \(f^{-1}(x) = -\sqrt{x}\) and substitute it into \(f(x)\). This gives \(f(f^{-1}(x)) = f(-\sqrt{x}) = (-\sqrt{x})^2 = x\). This confirms that \(f(f^{-1}(x)) = x\), verifying the first part of the identity function.

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

Take \(f(x) = x^{2}\), and substitute it into the inverse function \(f^{-1}(x)\). This gives \(f^{-1}(f(x)) = f^{-1}(x^{2}) = - \sqrt{x^{2}} = -|x| = x\), since we are considering \(x \leq 0\). This confirms that \(f^{-1}(f(x)) = x\), verifying the second part of the identity function.