To find the inverse function of

$\displaystyle y=3x+5$

we replace the $\displaystyle y$ with $\displaystyle x$ and vice versa.

So

$\displaystyle x=3y+5$

Now solve for $\displaystyle y$

$\displaystyle x-5=3y$

$\displaystyle 3y=x-5$

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

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

To find the inverse function, first replace $\displaystyle f(x)$ with $\displaystyle y$:

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

Now replace each $\displaystyle y$ with an $\displaystyle x$ and each $\displaystyle x$ with a $\displaystyle y$:

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

Solve the above equation for $\displaystyle y$:

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

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

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

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

To find the inverse function, first replace $\displaystyle g(x)$ with $\displaystyle y$:

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

Now replace each $\displaystyle y$ with an $\displaystyle x$ and each $\displaystyle x$ with a $\displaystyle y$:

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

Solve the above equation for $\displaystyle y$:

$\displaystyle x(2y-4)=y+3$

$\displaystyle 2xy-4x=y+3$

$\displaystyle 2xy-y=4x+3$

$\displaystyle (2x-1)y=4x+3$

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

Replace $\displaystyle y$ with $\displaystyle g^{-1}(x)$. This is the inverse function:

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

To find the inverse of $\displaystyle y=2x^2+1$, interchange the $\displaystyle x$ and $\displaystyle y$ terms and solve for $\displaystyle y$.

$\displaystyle y=2x^2+1$

$\displaystyle x=2y^2+1$

$\displaystyle x-1=2y^2$

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

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

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

So $\displaystyle (2,3)$ $\displaystyle \rightarrow (3,2)$

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

$\displaystyle y=4x+10$

After switching the variables, we have the following:

$\displaystyle x=4y+10$

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

$\displaystyle x-10=4y+10-10$

$\displaystyle x-10=4y$

Divide both sides of the equation by 4.

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

Rearrange and solve.

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

In order to find the inverse, switch the x and y variables in the function then solve for y.

$\displaystyle 3x+1=y$

Switching variables we get,

 $\displaystyle 3y+1=x$.

 Then solving for y to get our final answer.

$\displaystyle 3y=x-1$

$\displaystyle y=\frac{x-1}{3}$

First, switch the variables making $\displaystyle y=x^2$ into $\displaystyle x=y^{2}$.

Then solve for y by taking the square root of both sides.

$\displaystyle y^2=x$

$\displaystyle \sqrt {y^2}=\pm \sqrt x$

$\displaystyle y=\pm \sqrt x$

To find the inverse in this case, we need to switch our x and y variables and then solve for y.

Therefore,

$\displaystyle y=\sqrt{\ln(x)+2}$ becomes,

$\displaystyle x=\sqrt{\ln(y)+2}$

To solve for y we square both sides to get rid of the sqaure root.

$\displaystyle x^2=\ln(y)+2$

We then subtract 2 from both sides and take the exponenetial of each side, leaving us with the final answer.

$\displaystyle x^2-2=\ln(y)$

$\displaystyle e^{x^2-2}=e^{\ln(y)}$

$\displaystyle e^x^{^{2}-2}=y$