Determine and verify inverse functions

An inverse function reverses the input-output relationship of a one-to-one function, exchanging its domain and range; it can be determined algebraically by interchanging $x$ and $y$ and solving, with domain restrictions when necessary, such as for a quadratic. Inverses are verified through $f(f^{-1}(x))=x$ and $f^{-1}(f(x))=x$ on appropriate domains, and their graphs reflect across $y=x$ ; this is distinct from taking a reciprocal. The focus is on standard algebraic functions rather than abstract inverse relations or advanced generalized cases.

Detailed Explanation: 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:

Write $f(x)$ as $y$ .
Interchange $x$ and $y$ .
Solve for $y$ .
Rename the result $f^{-1}(x)$ .
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 $y=x$ .

Learn by doing: Determine and verify inverse functions

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Function Inverse - Two Functions to Is Inverse

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