Skill: Determine and verify inverse functions

An inverse function reverses the input-output process of a function. If $f(a)=b$, then $f^{-1}(b)=a$. To find an inverse algebraically:

1. Write $f(x)$ as $y$.
2. Interchange $x$ and $y$.
3. Solve for $y$.
4. Rename the result $f^{-1}(x)$.
5. Verify using composition.

A function must be one-to-one to have an inverse function. This means each output comes from only one input. For a quadratic, we usually restrict its domain so that it is one-to-one.

Worked example

Find and verify the inverse of

$$f(x)=(x-2)^2+1,\qquad x\ge 2.$$

The restriction $x\ge 2$ uses only the right half of the parabola, making the function one-to-one.

1. Replace $f(x)$ with $y$

$$y=(x-2)^2+1$$

2. Interchange $x$ and $y$

$$x=(y-2)^2+1$$

3. Solve for $y$

Subtract $1$:

$$x-1=(y-2)^2$$

Take the square root:

$$\sqrt{x-1}= \vert y-2 \vert$$

Because the original domain is $x\ge 2$, the corresponding outputs satisfy $y\ge 1$. Therefore, $y-2\ge -1$ is not enough by itself to choose the branch; more directly, the inverse must return values in the original domain $[2,\infty)$. Thus we choose the positive branch:

$$y-2=\sqrt{x-1}$$

So,

$$y=2+\sqrt{x-1}$$

Therefore,

$$\boxed{f^{-1}(x)=2+\sqrt{x-1}},\qquad x\ge 1.$$

The inverse’s domain is the original function’s range, $x\ge 1$. Its range is the original function’s domain, $y\ge 2$.

Verify the inverse

First check $f(f^{-1}(x))=x$:

$$f\left(f^{-1}(x)\right) =f\left(2+\sqrt{x-1}\right)$$ $$=\left((2+\sqrt{x-1})-2\right)^2+1$$ $$=\left(\sqrt{x-1}\right)^2+1 =x.$$

This is valid for $x\ge 1$.

Now check $f^{-1}(f(x))=x$:

$$f^{-1}(f(x)) =2+\sqrt{(x-2)^2+1-1}$$ $$=2+\sqrt{(x-2)^2} =2+ \vert x-2 \vert .$$

Since the original domain is $x\ge 2$, $\vert x-2 \vert =x-2$. Therefore,

$$f^{-1}(f(x))=2+(x-2)=x.$$

This is valid for $x\ge 2$.

Thus, the inverse is

$$\boxed{f^{-1}(x)=2+\sqrt{x-1}}.$$

Remember that an inverse is not the reciprocal $\frac{1}{f(x)}$. The graphs of $f$ and $f^{-1}$ are reflections of each other across the line $y=x$.