Pre-Calculus - Find the Inverse of a Function

Find the inverse of .

$y = \frac{x+1}{10}$

$y = \frac{x-1}{10}$

$x = \frac{y - 1}{10}$

Explanation

To find the inverse of the function, we switch the switch the and variables in the function.

Switching and gives

Then, solving for gives our answer:

$\frac{x-1}{10} = y$

Find the inverse of the follow function:

$y=\pm\sqrt{\frac{x+9}{3}}$

$y=\sqrt{\frac{x-9}{3}}$

$y=\frac{1}{3x^2+9}$

$y=\frac{1}{3x^2-9}$

Explanation

To find the inverse, substitute all x's for y's and all y's for x's and then solve for y.

$\frac{x+9}{3}=y^2$

$y=\pm\sqrt{\frac{x+9}{3}}$

Find the inverse function of .

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

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

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

None of the other answers.

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

Explanation

To find the inverse you must reverse the variables and solve for y.

Reverse the variables:

Solve for y:

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

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

Are these two function inverses? $f(x)=\sqrt{x-4}$ and .

Yes

Cannot tell

F(x) does not have an inverse.

G(x) does not have an inverse.

Explanation

One can ascertain if two functions have an inverse by finding the composition of both functions in turn. Each composition should equal x if the functions are indeed inverses of each other.

$f(g(x))=\sqrt{x^2+4-4}=x$

$g(f(x))=(\sqrt{x-4})^2+4=x$

The functions are inverses of each other.

Determine the inverse function, given

$y^{-1}=\sqrt[3]{x-2}$

$y^{-1}=x^3-2$

$y^{-1}=(x-2)^3$

Explanation

In order to find the inverse function we

switch the variables and
solve for the new variable

For the function

...

$y=\sqrt[3]{x-2}$

Hence, the inverse function is

$y^{-1}=\sqrt[3]{x-2}$

If $f(x)=\frac{1+3x}{2-x}$ , find $f^{-1}(1)$ .

$\frac{1}{4}$

$\frac{2}{3}$

$-\frac{3}{4}$

$\frac{1}{3}$

$\frac{4}{3}$

Explanation

Set , thus $y=\frac{1+3x}{2-x}$ .

Now switch with .

So now,

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

Simplify to isolate by itself.

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

Therefore,

$\ y(3+x)=2x-1\ \ y=\frac{2x-1}{3+x}$ .

Now substitute with $f^{-1}(x)$ ,

$f^{-1}(x)=\frac{2x-1}{3+x}$ , and

$f^{-1}(1)=\frac{2(1)-1}{3+1}=\frac{1}{4}$ .

If $f(x)=-\frac{3x}{5}+11$ , what is its inverse function, $f^{-1}(x)$ ?

$f^{-1}(x)=-\frac{5x}{3}+\frac{55}{3}$

$f^{-1}(x)=-\frac{5x}{3}-\frac{55}{3}$

$f^{-1}(x)=\frac{5x}{3}+\frac{55}{3}$

$f^{-1}(x)=-\frac{3x}{5}+\frac{33}{5}$

$f^{-1}(x)=-\frac{3x}{5}-\frac{33}{5}$

Explanation

We begin by taking $f(x)=-\frac{3x}{5}+11$ and changing the to a , giving us $y=-\frac{3x}{5}+11$ .

Next, we switch all of our and , giving us $x=-\frac{3y}{5}+11$ .

Finally, we solve for by subtracting from each side, multiplying each side by , and dividing each side by , leaving us with,

$y=-\frac{5x}{3}+\frac{55}{3}$ .

Find the inverse of this function:

$g^{-1}(x)= \frac{x^{0.2}}{2}$

$g^{-1}(x)= \frac{1}{2}x^5$

$g^{-1}(x)=5 \sqrt x$

$g^{-1}(x)=\left({\frac{1}{2}x}\right)^{0.2}$

$g^{-1}(x)=\sqrt[5]{2x}$

Explanation

In order to have the inverse of a function, the new function must perform the inverse opperations in the opposite order. One way to ensure that is true is to consider the case of , switch x and y, then solve for y.

in this case becomes .

Our first step in solving is to take the reciprocal power on each side.

The reciprocal of 5 is $\frac{1}{5}=0.2$ , so we'll take both sides to the power of 0.2:

$x^{0.2}= 2y$ Now divide by 2:

$\frac{x^{0.2}}{2}=y$

Note that the answer $g^{-1}(x)=\left({\frac{1}{2}x}\right)^{0.2}$ has the correct inverse opperations, it is just in the wrong order - first you divide by 2, then you take x to the power of 0.2.