Find the Inverse of a Relation - Pre-Calculus

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What is the inverse function of

?

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To find the inverse function of

we replace the with and vice versa.

So

Now solve for

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

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

What is the inverse of $\left($\frac{1}{2}$,3\right)$ ?

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When trying to find the inverse of a point, switch the x and y values.

So, $\left($\frac{1}{2}$,3\right)\rightarrow \left(3, $\frac{1}{2}$ \right )$

Find the inverse of the following function:

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In order to find the inverse of the function, we need to switch the x- and y-variables.

After switching the variables, we have the following:

Now solve for the y-variable. Start by subtracting 10 from both sides of the equation.

Divide both sides of the equation by 4.

$\frac{x-10}{4}$=\frac{4y}{4}$

Rearrange and solve.

$y=\frac{x-10}{4}$$

Find the inverse of the function.

$f(x)= $$\sin{\frac{x^2$}{10}$}$

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To find the inverse function, first replace with :

$y= $\sin{ $\frac{x^2$}{10}$ }$

Now replace each with an and each with a :

$x= $\sin{ $\frac{y^2$}{10}$ }$

Solve the above equation for :

$\arc$\sin{x}$= $$\frac{y^2$}{10}$$

$$\sqrt{10 \arcsin{x}$} = y$

Replace with . This is the inverse function:

Find the inverse of the function.

$g(x)= $\frac{x+3}{2x-4}$$

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To find the inverse function, first replace with :

$y= $\frac{x+3}{2x-4}$$

Now replace each with an and each with a :

$x= $\frac{y+3}{2y-4}$$

Solve the above equation for :

$y= $\frac{4x+3}{2x-1}$$

Replace with . This is the inverse function:

Find the inverse of the function .

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To find the inverse of , interchange the and terms and solve for .

$f^$-^1$(x)=\pm $\sqrt{\frac{x-1}{2}$}$

What point is the inverse of the ?

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When trying to find the inverse of a point, switch the x and y values.

So $\rightarrow (3,2)$

What is the inverse function of

?

Tap to see back →

To find the inverse function of

we replace the with and vice versa.

So

Now solve for

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

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

What is the inverse of $\left($\frac{1}{2}$,3\right)$ ?

Tap to see back →

When trying to find the inverse of a point, switch the x and y values.

So, $\left($\frac{1}{2}$,3\right)\rightarrow \left(3, $\frac{1}{2}$ \right )$

Find the inverse of the following function:

Tap to see back →

In order to find the inverse of the function, we need to switch the x- and y-variables.

After switching the variables, we have the following:

Now solve for the y-variable. Start by subtracting 10 from both sides of the equation.

Divide both sides of the equation by 4.

$\frac{x-10}{4}$=\frac{4y}{4}$

Rearrange and solve.

$y=\frac{x-10}{4}$$

Find the inverse of the function.

$f(x)= $$\sin{\frac{x^2$}{10}$}$

Tap to see back →

To find the inverse function, first replace with :

$y= $\sin{ $\frac{x^2$}{10}$ }$

Now replace each with an and each with a :

$x= $\sin{ $\frac{y^2$}{10}$ }$

Solve the above equation for :

$\arc$\sin{x}$= $$\frac{y^2$}{10}$$

$$\sqrt{10 \arcsin{x}$} = y$

Replace with . This is the inverse function:

Find the inverse of the function.

$g(x)= $\frac{x+3}{2x-4}$$

Tap to see back →

To find the inverse function, first replace with :

$y= $\frac{x+3}{2x-4}$$

Now replace each with an and each with a :

$x= $\frac{y+3}{2y-4}$$

Solve the above equation for :

$y= $\frac{4x+3}{2x-1}$$

Replace with . This is the inverse function:

Find the inverse of the function .

Tap to see back →

To find the inverse of , interchange the and terms and solve for .

$f^$-^1$(x)=\pm $\sqrt{\frac{x-1}{2}$}$

What point is the inverse of the ?

Tap to see back →

When trying to find the inverse of a point, switch the x and y values.

So $\rightarrow (3,2)$

What is the inverse function of

?

Tap to see back →

To find the inverse function of

we replace the with and vice versa.

So

Now solve for

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

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

Find the inverse of the following function:

Tap to see back →

In order to find the inverse of the function, we need to switch the x- and y-variables.

After switching the variables, we have the following:

Now solve for the y-variable. Start by subtracting 10 from both sides of the equation.

Divide both sides of the equation by 4.

$\frac{x-10}{4}$=\frac{4y}{4}$

Rearrange and solve.

$y=\frac{x-10}{4}$$

Find the inverse of the function.

$f(x)= $$\sin{\frac{x^2$}{10}$}$

Tap to see back →

To find the inverse function, first replace with :

$y= $\sin{ $\frac{x^2$}{10}$ }$

Now replace each with an and each with a :

$x= $\sin{ $\frac{y^2$}{10}$ }$

Solve the above equation for :

$\arc$\sin{x}$= $$\frac{y^2$}{10}$$

$$\sqrt{10 \arcsin{x}$} = y$

Replace with . This is the inverse function:

Find the inverse of the function.

$g(x)= $\frac{x+3}{2x-4}$$

Tap to see back →

To find the inverse function, first replace with :

$y= $\frac{x+3}{2x-4}$$

Now replace each with an and each with a :

$x= $\frac{y+3}{2y-4}$$

Solve the above equation for :

$y= $\frac{4x+3}{2x-1}$$

Replace with . This is the inverse function:

Find the inverse of the function .

Tap to see back →

To find the inverse of , interchange the and terms and solve for .

$f^$-^1$(x)=\pm $\sqrt{\frac{x-1}{2}$}$

What point is the inverse of the ?

Tap to see back →

When trying to find the inverse of a point, switch the x and y values.

So $\rightarrow (3,2)$