Question

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

Short answer

The inverse function of \(f(x) = -(x - 2)^2\) is \(f^{-1}(x) = 2 - \sqrt{-x}\). The function \(f(x)\) and its inverse have both been verified as \(\left(f \circ f^{-1}\right)(x)=x\) and \(\left(f^{-1} \circ f\right)(x)=x\) respectively.

Step-by-step solution

Find the inverse of the function

To find the inverse function \(f^{-1}(x)\), we start by replacing \(f(x)\) with \(y\), thus the equation becomes \(y = -(x - 2)^2\). Then, we switch the roles of 'x' and 'y', resulting in \(x = - (y - 2)^2\). Solving this equation, we isolate 'y': first we remove the negative on the right hand side which gives \(x = -1 \cdot ((y - 2)^2)\), that is, \(x = -(y - 2)^2\). Eventually, solving for 'y', we obtain \(y = 2 - \sqrt{-x}\). This brings us to our inverse function, \(f^{-1}(x) = 2 - \sqrt{-x}\).

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

To verify this, we substitute \(f^{-1}(x)\) into \(f(x)\): \(f(f^{-1}(x)) = -((2 - \sqrt{-x}) - 2)^2. This simplifies to \(f(f^{-1}(x)) = -(-\sqrt{-x})^2\), which ultimately simplifies to \(f(f^{-1}(x)) = -x\). This verifies that \(\left(f \circ f^{-1}\right)(x)=x\).

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

To verify this, we substitute \(f(x)\) into \(f^{-1}(x)\): \(f^{-1}(f(x)) = 2 - \sqrt{-(-(x - 2)^2)}\). This simplifies to \(f^{-1}(f(x)) = 2 - (x - 2) = x\), which verifies that \(\left(f^{-1} \circ f\right)(x)=x\).

Key concepts

Finding Inverse Functions

Understanding how to find the inverse of a function is essential in calculus, as it reveals a unique relationship where two functions, in a sense, 'undo' one another. To find the inverse function, often denoted as \(f^{-1}(x)\), we perform a series of steps. Initially, we set \(f(x)\) equal to the variable \(y\), creating an equation where \(y\) depends on \(x\). Then, to find the inverse, we swap the roles of \(x\) and \(y\) and solve for the new \(y\), which gives us the inverse function.

In our exercise, after replacing \(f(x)\) with \(y\) and switching \(x\) with \(y\), we manipulate the resulting equation to solve for the new \(y\). This process involves basic algebraic steps such as isolating terms and taking square roots while respecting the domain restriction of \(x \leq 2\) to ensure we have a function. Our resulting function \(f^{-1}(x) = 2 - \sqrt{-x}\) is the inverse of the original function \(f(x)\).

Composition of Functions

To deepen our understanding, it's important to grasp the concept of the composition of functions. Composing functions involves combining two functions in a way that the output of one function becomes the input of the other. Notationally, this is written as \((f \circ g)(x)\), which means \(f(g(x))\).

Practical Application in Finding Inverses

When we find the inverse of a function, we essentially search for a function that, when composed with the original, will yield the identity function \(x\). In the given exercise, the composition \((f \circ f^{-1})(x)\) and \((f^{-1} \circ f)(x)\) should both result in \(x\). This is the heart of verifying that we have indeed found a true inverse function.

Verification of Inverse Functions

Verifying that two functions are inverses of each other is a crucial step. It confirms that each function correctly 'undoes' the operation of the other. The main verification method is to show that their compositions give us the identity function. Essentially, we need to demonstrate that \((f \circ f^{-1})(x) = x\) and \((f^{-1} \circ f)(x) = x\).

This verification acts as a proof of our work. For the given function \(f(x) = -(x-2)^2\), with its inverse \(f^{-1}(x) = 2 - \sqrt{-x}\), we substitute one function into the other, simplify, and if our algebra is correct, we'll arrive at the simple function \(x\). In this way, our exercise does not only show us how to find an inverse, but also why the steps we've taken are accurate reflections of the inverse relationship.