Functions

Lesson · Beginner

Algebra

A function assigns exactly one output to each input.

f:A\to B

y=f(x),

then x is the input and y is the output.

The domain is the set of all allowed inputs.

\operatorname{Domain}(f)=\{x:f(x)\text{ is defined}\}

The range is the set of all outputs actually produced by the function.

\operatorname{Range}(f)=\{f(x):x\in\operatorname{Domain}(f)\}

For example,

f(x)=\sqrt{x-2}

has

\operatorname{Domain}(f)=[2,\infty)

and

\operatorname{Range}(f)=[0,\infty).

\operatorname{Domain}(f)=\{1,2,3,4\}

\operatorname{Range}(f)=\{a,b,d\}

A function is one-to-one (injective) if different inputs always give different outputs.

x_1\ne x_2\quad\Longrightarrow\quad f(x_1)\ne f(x_2)

Equivalently,

f(x_1)=f(x_2)\quad\Longrightarrow\quad x_1=x_2.

A function is one-to-one exactly when every horizontal line intersects its graph at most once.

This is called the horizontal line test.

If f is one-to-one, we can reverse the correspondence between inputs and outputs.

The resulting function is called the inverse function and is denoted by

f^{-1}.

f(a)=b,

then

f^{-1}(b)=a.

Thus,

\operatorname{Domain}(f^{-1})=\operatorname{Range}(f)

and

\operatorname{Range}(f^{-1})=\operatorname{Domain}(f).

To find $f^{-1}$ :

1. $y=f(x)$

2. Interchange $x$ and $y$

3. Solve for $y$

4. Replace $y$ by $f^{-1}(x)$

For example, if

f(x)=3x+5,

then

y=3x+5

Interchange x and y:

x=3y+5

y=\frac{x-5}{3}.

Therefore,

f^{-1}(x)=\frac{x-5}{3}

A function and its inverse undo each other:

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

and

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

whenever the expressions are defined.

The graphs of $f$ and $f^{-1}$ are reflections of each other across the line

y=x

because if

f(a)=b,

then

f^{-1}(b)=a.

Thus, if $(a,b)$ is a point on the graph of $f$ , then $(b,a)$ is a point on the graph of $f^{-1}$ .