 # Inverse Function Calculator + Online Solver With Free Steps

The Inverse Function Calculator finds the inverse function g(y) if it exists for the given function f(x). If the inverse function does not exist, the calculator looks for an inverse relation. The input function must be a function of only x. If x is not present in the input, the calculator will not work.

The calculator does not support finding the inverse of multi-variable functions of the form f(x1, x2, x3, … , xn) for all n variables. If you enter such a function, it considers all variables other than x as constants, and solves only for f(x). ## What Is the Inverse Function Calculator?

The Inverse Function Calculator is an online tool that calculates the inverse function or relation $\mathbf{g(y)}$ for the input function $\mathbf{f(x)}$ such that feeding the output of $\mathbf{f(x)}$ to $\mathbf{g(y)}$ undoes the effect of $\mathbf{f(x)}$.

The calculator interface consists of a single text box labeled “The inverse function of.” In this, you simply enter the input expression as a function of x. After that, you just submit it for calculation.

## How To Use the Inverse Function Calculator?

You can use the Inverse Function Calculator by entering the function whose inverse you want to find. The step-by-step guidelines are below.

For example, suppose we want to find the inverse of f(x)=3x-2.

### Step 1

Enter the function into the text box. For our case, we type “3x-2” here. We could also enter “y=3x-2” as it means the same thing.

### Step 2

Click the Submit button to calculate the inverse function.

### Results

The results open up in a new pop-up window. For our example, the inverse function is:

$\frac{x+2}{3}$

The result’s variable x is not to be confused with the variable x in the input function f(x). In the terminology used to describe the calculator so far, the x in the results is equivalent to y in g(y) and represents the output value of the input function.

For example, in our case:

f(x=10) = 3(10)-2 = 28

Now if we put x = 28 in the calculator’s output inverse function:

$\frac{28+2}{3} = \frac{30}{3} = 10$

That is the original value fed to f(x).

## How Does the Inverse Function Calculator Work?

The Inverse Function Calculator works by using the variable/coordinate swapping method to find the inverse function. Essentially, given that ‘*’ is any defined operator:

f(x) = terms with x  * other terms with constants

Put f(x)=y. This represents the value of the function at x. Our equation is then:

y = terms with x * other terms with constants *{(1)}

Now swap the variables x and y:

x = terms with y * other terms with constants

And solve for y in terms of x to get the inverse mapping. You can get the same result by solving for x in equation (1), but the variable swap keeps things neat by keeping the usual function nomenclature (x is the input, y is the output).

You can see that the technique uses the known output of the function to find the input given that we know the function itself. Thus, the resulting inverse function g(x) is also in terms of x, but remember that we swapped the variables, so this x represents the output of the first function (y), not the input.

### Inverse Function Definition

The function g(y) is the inverse function of f(x) only if:

$y = f(x) \iff x = g(y) \, \Rightarrow \, g(f(x)) = x \,\, \text{and} \,\, f(g(y)) = y$

In other words, if f: X to Y, then g: Y to X which can be read as: if applying f to a value x gives the output y, then applying the inverse function g to y would give back the original input x, essentially undoing the effect of f(x).

Note that g(f(x)) = g $\circ$ f is the composition of the inverse function with the original function. Often the inverse function g(y) is notated as $f^{-1}(y)$ such that if f: X to Y, then:

$f^{-1}(f(x)) = x \,\, \text{and} \,\, f \left( f^{-1}(y) \right) = x$

It follows that the inverse of an inverse function g(y) is the original function y = f(x):

$f^{-1} \left( f^{-1}(y) \right) = y \, \Rightarrow \, g(g(y)) = y$

### Existence of the Inverse

Note that g(y) might not necessarily be a function (one input, one output) but a relation (one input to multiple outputs). Generally, this happens when the input function is bijective or many-to-one (that is, it maps different inputs to the same output). In such a case, the exact input is irrecoverable and the inverse function does not exist.

It is possible, however, that an inverse relation exists. You can tell if the calculator output is an inverse relation if it shows more than one output or a ‘$\pm$’ sign.

Examples of functions that do not have an inverse function are $f(x) = x^2$ and f(x) = |x|. Because the output of the functions has the same output (value of y) for multiple inputs (values of x), the inverse does not uniquely return x as it returns multiple values of x that satisfy the relation.

### Horizontal Line Test

The horizontal line test is sometimes used to check if the input function is bijective. If you can draw a horizontal line that intersects the function’s graph at more than one point, then that function is many-to-one, and its inverse is at best a relation.

## Solved Examples

Here are some examples to help us understand the topic further.

### Example 1

Find the inverse function for the function:

f(x)= 3x-2

### Solution

Let:

f(x) = y  $\Rightarrow$ y=3x-2

Now swap x and y so that now we have the original input x as a function of the output value y:

x = 3y-2

Solving for y:

$x + 2 = 3y \, \Rightarrow \, y = \frac{x+2}{3}$

That is the required inverse function. The calculator also shows this result.

### Example 2

For the function

$f(x) = 10\ln \left( \frac{1}{1+x} \right)$

Find the inverse and classify it as a function or a relation. Verify this for the input x=10.

### Solution

Using the same substitution method as in Example 1, we first re-write:

$y = f(x) \, \Rightarrow \, y = 10\ln \left( \frac{1}{1+x} \right)$

Now swap the variables and solve for y:

$x = 10\ln \left( \frac{1}{1+y} \right)$

$\frac{1}{10} \cdot x = \ln \left( \frac{1}{1+y} \right)$

$\frac{x}{10} = \ln \left( \frac{1}{1+y} \right) \, \Rightarrow \, 0.1x = \ln \left( \frac{1}{1+y} \right)$

Taking the inverse of the natural log on both sides:

$\ln^{-1} \left( 0.1x \right) = \ln^{-1} \left\{ \ln \left( \frac{1}{1+y} \right) \right\}$

Given that:

$\because \ln^{-1}(a) = e^a \,\, \text{and} \,\, \ln^{-1}\{\ln(x)\} = x$

$\Rightarrow e^{ 0.1x } = \frac{1}{1+y}$

Multiplying both sides by $(1+y)$:

$(1+y) \left( e^{ 0.1x } \right) = 1$

Dividing both sides by $e^{\left (0.1x \right)}$:

$1+y= \frac{1}{e^{ 0.1x}}$

$\Rightarrow y = \frac{1}{e^{ 0.1x}}-1$

Which can be re-arranged as:

$y = \frac{1-e^{0.1x}}{e^{ 0.1x}}$

$y = -e^{-0.1x} \left( e^{ 0.1x}-1 \right)$

That is the result shown by the calculator (in fraction form).

Verifying for x=10:

$f(x=10) = y = 10\ln \left( \frac{1}{1+10} \right) \, \Rightarrow \, y \approx -23.97895$

$g(y=-23.97895) = x = -e^{-0.1y} \left( e^{ 0.1y}-1 \right) \, \Rightarrow \, y = 9.99999 \approx 10$

That is correct.

### Example 3

Given the function:

$f(x) = 30x^2-15x+x\ln(10)$

Find the inverse function if it exists. Else, find the inverse relation and explain why it is a relation.

### Solution

The function is quadratic. Its graph will be a parabola, so we can see that it will not have an inverse function because a horizontal line will always intersect a parabola at more than one point. Because it is bijective (many-to-one), it is not invertible.

However, we could try to find the inverse relation using the same technique of variable swapping used earlier.

$y = 30x^2-15x+x\ln(10)$

$x = 30y^2-15y+y\ln(10)$

Given that $x$ is the value of the function, we treat it as a constant. Re-arranging:

$\Rightarrow 30y^2+\left( -15+\ln 10 \right)y-x = 0$

Since this is a quadratic function with a=30, b=15-ln(10) and c=x, we use the quadratic formula to solve for y:

$y_1,\, y_2 = \frac{-b \pm \sqrt{b^2-4ac}}{2a}$

Let $\mathbf{y}=\{y_1,\, y_2\}$, then:

$\mathbf{y} = \frac{15-\ln10 \pm \sqrt{\left(-15+\ln10 \right)^2-4(30)(x)}}{2(30)}$

$\mathbf{y} = \frac{15-\ln10 \pm \sqrt{225-30\ln(10)+\ln^2(10)-120x}}{60}$

Which gives us the inverse relation. The two possible solutions are then:

$g(y=y_1) = \frac{15-\ln10-\sqrt{\left(-15+\ln10 \right)^2-4(30)(x)}}{2(30)}$

$g(y=y_2) = \frac{15-\ln10 + \sqrt{\left(-15+\ln10 \right)^2-4(30)(x)}}{2(30)}$

Clearly, the same value of y = f(x) will give two solutions for x = g(y) so our original function f(x) is not bijective, and the inverse mapping is a relation, not a function.