Inverse Functions

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Pre-Calculus › Inverse Functions

Questions 1 - 10

1

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:

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

2

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:

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

3

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.

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

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

4

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.

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

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

5

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:

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

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

6

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:

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

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

7

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

Yes

No

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.

8

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

Yes

No

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.

9

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.

So

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

Therefore,

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

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

so

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

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

10

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.

So

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

Therefore,

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

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

so

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

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

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