Find the Inverse of a Relation

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Pre-Calculus › Find the Inverse of a Relation

Questions 1 - 7

1

Find the inverse of the function .

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

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

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

Explanation

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

$\frac{x-1}{2}=y^2$

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

2

What point is the inverse of the ?

Explanation

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

So $\rightarrow (3,2)$

3

Find the inverse of the following function:

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

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

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

Explanation

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.

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

4

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

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

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

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

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

Explanation

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

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

5

Find the inverse of the function.

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

$g^{-1}(x)= \frac{4x+3}{2x-1}$

$g^{-1}(x)= \frac{3x+4}{x-2}$

$g^{-1}(x)= \frac{x+3}{-4x+1}$

$g^{-1}(x)= \frac{x+4}{2-x}$

Explanation

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 $g^{-1}(x)$ . This is the inverse function:

$g^{-1}(x)= \frac{4x+3}{2x-1}$

6

Find the inverse of the function.

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

$f^{-1}(x)=\sqrt{ 10 \arcsin{x} }$

$f^{-1}(x)=\arcsin{ \frac{x^2}{10} }$

$f^{-1}(x)=10 \sqrt{ \arcsin{y} }$

$f^{-1}(x)=\sqrt{ 10 \cos{x} }$

Explanation

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 :

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

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

Replace with $f^{-1}(x)$ . This is the inverse function:

$f^{-1}(x)=\sqrt{ 10 \arcsin{x} }$

7

What is the inverse function of

?

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

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

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

Explanation

To find the inverse function of

we replace the with and vice versa.

So

Now solve for

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

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

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