Inverse of a Function

We already know that a function is a rule that allows finding for every value of x{x} the corresponding value of y{y}. But what if we are given a rule, i.e. a function, and the value of y{y} and we want to find the corresponding value of x{x}?

In other words, we want to find a function that will express x{x} in terms of y{y}. Such a function is called an inverse function and is denoted by f1{{f}}^{{-{1}}}.

If y=f(x){y}={f{{\left({x}\right)}}}, the inverse function is x=f1(y){x}={{f}}^{{-{1}}}{\left({y}\right)}.

For example, if y=f(x)=x3{y}={f{{\left({x}\right)}}}={{x}}^{{3}}, it can be stated that f1(27)=3{{f}}^{{-{1}}}{\left({27}\right)}={3}, because f(3)=27{f{{\left({3}\right)}}}={27}.

Not all functions have a unique inverse. Let's take the function f(x)=x2{f{{\left({x}\right)}}}={{x}}^{{2}}. Since f(2)=4{f{{\left(-{2}\right)}}}={4} and f(2)=4{f{{\left({2}\right)}}}={4}, it can be concluded that f1(4){{f}}^{{-{1}}}{\left({4}\right)} is undefined, because there are 2 values that correspond to 4{4}, namely 2{2} and 2-{2}. Therefore, the inverse of y=x2{y}={{x}}^{{2}} is a multi-valued function. In this case, we will say that the function doesn't have an inverse, because the inverse is a multi-valued function, which we don't consider to be a function.

So, a function will have an inverse only when for different values of x{x} we have different values of y{y}.

Definition: a function f{f{}} is called one-to-one (or injective), if it never takes the same value twice, i.e. f(x1)f(x2){f{{\left({x}_{{1}}\right)}}}\ne{f{{\left({x}_{{2}}\right)}}} whenever x1x2{x}_{{1}}\ne{x}_{{2}}.

one to one functions

One-to-one functions are precisely the functions that possess inverse functions.

From this definition, it follows that y=x2{y}={{x}}^{{2}} is not a one-to-one function, because y(2)=y(2)=4{y}{\left({2}\right)}={y}{\left(-{2}\right)}={4}; so, different values have the same output.

On the other hand, y=x3{y}={{x}}^{{3}} is a one-to-one function, because x13x23{{x}_{{1}}^{{3}}}\ne{{x}_{{2}}^{{3}}} whenever x1x2{x}_{{1}}\ne{x}_{{2}} (different values can't have the same cube). Thus, y=x3{y}={{x}}^{{3}} has an inverse.

Graphically, we can determine whether a function is one-to-one using the horizontal line test.

Horizontal line test. A function is one-to-one if and only if no horizontal line intersects its graph more than once.

horizontal line test

Now, let's give a more formal definition of the inverse function.

Definition. Let f{f{}} be a one-to-one function with the domain X{X} and the range Y{Y}. Then, its inverse function f1{{f}}^{{-{1}}} has the domain Y{Y} and the range X{X} and is defined as follows: f1(y)=x  f(x)=y{{f}}^{{-{1}}}{\left({y}\right)}={x}\ \Leftrightarrow\ {f{{\left({x}\right)}}}={y}.

Note that the domain of f{f{}} equals the range of f1{{f}}^{{-{1}}} and the range of f{f{}} equals the domain of f1{{f}}^{{-{1}}}.

When we want to concentrate on the inverse function, we will usually write y=f1(x){y}={{f}}^{{-{1}}}{\left({x}\right)}, i.e. we will interchange x{x} and y{y}, because we traditionally use x{x} as an independent variable.

The inverse function means that if we apply f{f{}} to x{x}, we will obtain y{y}.

inverse function

If we now apply f1{{f}}^{{-{1}}} to y{y}, we will arrive back at x{x}. And vice versa: applying f1{{f}}^{{-{1}}} to y{y} and then applying f{f{}} to the result will bring us back to y{y}.

As a result, we can say that a function and its inverse cancel the influence of each other.

Thus, there are two properties of inverse functions:

  1. f1(f(x))=x{{f}}^{{-{1}}}{\left({f{{\left({x}\right)}}}\right)}={x} for any x{x} from X{X}.
  2. f(f1(x))=x{f{{\left({{f}}^{{-{1}}}{\left({x}\right)}\right)}}}={x} for any x{x} from Y{Y}.

Example 1. Given f(1)=4{f{{\left({1}\right)}}}={4}, f(2)=3{f{{\left({2}\right)}}}={3}, f(5)=10{f{{\left({5}\right)}}}={10}, find f1(3){{f}}^{{-{1}}}{\left({3}\right)}, f1(4){{f}}^{{-{1}}}{\left({4}\right)}, f1(5){{f}}^{{-{1}}}{\left({5}\right)}.

f1(3)=2{{f}}^{{-{1}}}{\left({3}\right)}={2}, because f(2)=3{f{{\left({2}\right)}}}={3}.

f1(4)=1{{f}}^{{-{1}}}{\left({4}\right)}={1}, because f(1)=4{f{{\left({1}\right)}}}={4}.

f1(5){{f}}^{{-{1}}}{\left({5}\right)} is undefined, because we are not given the value of x{x} for which f(x)=5{f{{\left({x}\right)}}}={5}.

Now, let’s see how to compute inverse functions. Since we have y=f(x){y}={f{{\left({x}\right)}}}, we need to express x{x} in terms of y{y} to obtain f1{{f}}^{{-{1}}}.

Steps for finding an inverse:

  1. Write y=f(x){y}={f{{\left({x}\right)}}}.
  2. Express x{x} in terms of y{y} (if possible). You will get x=f1(y){x}={{f}}^{{-{1}}}{\left({y}\right)}.
  3. To obtain y=f1(x){y}={{f}}^{{-{1}}}{\left({x}\right)}, interchange x{x} and y{y}.

Actually, we can first perform step 3 and then step 2, i.e. interchange x{x} and y{y}, and then express y{y} in terms of x{x}.

Example 2. Find the inverse of y=x33{y}={{x}}^{{3}}-{3}.

  1. y=x33{y}={{x}}^{{3}}-{3}.
  2. x=y+33{x}={\sqrt[{{3}}]{{{y}+{3}}}}.
  3. y=x+33{y}={\sqrt[{{3}}]{{{x}+{3}}}} is the inverse.

Let's work another short example.

Example 3. Find the inverse of y=x+32x5{y}=\frac{{{x}+{3}}}{{{2}{x}-{5}}}.

  1. y=x+32x5{y}=\frac{{{x}+{3}}}{{{2}{x}-{5}}}.
  2. 2xy5y=x+3{2}{x}{y}-{5}{y}={x}+{3}, or x=5y+32y1{x}=\frac{{{5}{y}+{3}}}{{{2}{y}-{1}}}.
  3. y=5x+32x1{y}=\frac{{{5}{x}+{3}}}{{{2}{x}-{1}}} is the inverse.

To verify whether we found the inverse correctly, check if f(f1(x))=x{f{{\left({{f}}^{{-{1}}}{\left({x}\right)}\right)}}}={x} holds.

Here, f(x)=x+32x5{f{{\left({x}\right)}}}=\frac{{{x}+{3}}}{{{2}{x}-{5}}}, and f1(x)=5x+32x1{{f}}^{{-{1}}}{\left({x}\right)}=\frac{{{5}{x}+{3}}}{{{2}{x}-{1}}}; so, f(f1(x))=5x+32x1+325x+32x15=x{f{{\left({{f}}^{{-{1}}}{\left({x}\right)}\right)}}}=\frac{{\frac{{{5}{x}+{3}}}{{{2}{x}-{1}}}+{3}}}{{{2}\frac{{{5}{x}+{3}}}{{{2}{x}-{1}}}-{5}}}={x}.

Thus, we found the inverse correctly.

Interchanging the roles of x{x} and y{y} gives the opportunity to plot the graph of the inverse function based on the graph of the function y=f(x){y}={f{{\left({x}\right)}}}. Since the point (x0,y0){\left({x}_{{0}},{y}_{{0}}\right)} lies on the graph of f{f{}}, we have that (y0,x0){\left({y}_{{0}},{x}_{{0}}\right)} lies on the graph of f1{{f}}^{{-{1}}}. So, the graph of the inverse is obtained by reflecting the graph of the function about the line y=x{y}={x}.

Example 4. Sketch the graph of y=x1{y}=\sqrt{{{x}-{1}}} and its inverse.

The domain of y=x1{y}=\sqrt{{{x}-{1}}} is x10{x}-{1}\ge{0}, or x1{x}\ge{1}. The range is y0{y}\ge{0}.

graph of inverse

This means that the domain of the inverse is x0{x}\ge{0} and the range is y1{y}\ge{1}.

To find the equation of the inverse, we write x{x} in terms of y{y}: y2=x1{{y}}^{{2}}={x}-{1}, or x=y2+1{x}={{y}}^{{2}}+{1}. Now, interchange y{y} and x{x}: y=x2+1{y}={{x}}^{{2}}+{1}.

So, the inverse is y=x2+1{y}={{x}}^{{2}}+{1}.

Now, we draw the graph of the function y=x1{y}=\sqrt{{{x}-{1}}} on the interval [1,){\left[{1},\infty\right)} and then reflect it about the line y=x{y}={x}.