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Precalculus : Inverse Functions

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Example Question #1 : Find The Inverse Of A Relation

What is the inverse function of

y=3x+5 ?

Possible Answers:

y=3x+5

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

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

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

y=x-5

Correct answer:

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

Explanation:

To find the inverse function of

y=3x+5

we replace the with and vice versa.

x=3y+5

Now solve for

x-5=3y

3y=x-5

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

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

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Example Question #2 : Find The Inverse Of A Relation

Find the inverse of the function.

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

Possible Answers:

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

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

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

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

Correct answer:

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

Explanation:

To find the inverse function, first replace f(x) 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:

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

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Example Question #5 : Linear Algebra

Find the inverse of the function.

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

Possible Answers:

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

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

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

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

Correct answer:

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

Explanation:

To find the inverse function, first replace g(x) 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 :

x(2y-4)=y+3

2xy-4x=y+3

2xy-y=4x+3

(2x-1)y=4x+3

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

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

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

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Example Question #6 : Linear Algebra

Find the inverse of the function y=2x^2+1 .

Possible Answers:

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

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

f^-^1(x)=x+1

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

f^-^1(x)=-x+1

Correct answer:

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

Explanation:

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

y=2x^2+1

x=2y^2+1

x-1=2y^2

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

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

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Example Question #1 : Find The Inverse Of A Relation

What point is the inverse of the (2,3) ?

Possible Answers:

(3,2)

(-2,-3)

(2,-3)

(-2,3)

Correct answer:

(3,2)

Explanation:

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

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

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Example Question #1 : Find The Inverse Of A Relation

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

Possible Answers:

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

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

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

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

Correct answer:

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

Explanation:

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

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

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Example Question #5 : Find The Inverse Of A Relation

Find the inverse of the following function:

y=4x+10

Possible Answers:

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

y=10x-4

y=4x-10

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

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

Correct answer:

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

Explanation:

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

y=4x+10

After switching the variables, we have the following:

x=4y+10

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

x-10=4y+10-10

x-10=4y

Divide both sides of the equation by 4.

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

Rearrange and solve.

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

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Example Question #1 : Find The Inverse Of A Function

Find the inverse of,

3x+1=y .

Possible Answers:

$y=\frac{x+1}{3}$

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

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

x=3y+1

Correct answer:

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

Explanation:

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

3x+1=y

Switching variables we get,

3y+1=x .

Then solving for y to get our final answer.

3y=x-1

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

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Example Question #2 : Find The Inverse Of A Function

Find the inverse of,

$y=x^{2}$ .

Possible Answers:

$x=y^{2}$

$x=\pm \sqrt{y}$

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

$y=\pm\sqrt{x}$

Correct answer:

$y=\pm\sqrt{x}$

Explanation:

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

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

y^2=x

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

$y=\pm \sqrt x$

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Example Question #3 : Find The Inverse Of A Function

Find the inverse of the following equation.

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

Possible Answers:

$y=\pm x^2 -2$

y=x-2

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

y=e^x+2

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

Correct answer:

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

Explanation:

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

Therefore,

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

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

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

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

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

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

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

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

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Precalculus : Inverse Functions

Study concepts, example questions & explanations for Precalculus

All Precalculus Resources

Example Questions

Example Question #1 : Find The Inverse Of A Relation

Example Question #2 : Find The Inverse Of A Relation

Example Question #5 : Linear Algebra

Example Question #6 : Linear Algebra

Example Question #1 : Find The Inverse Of A Relation

Example Question #1 : Find The Inverse Of A Relation

Example Question #5 : Find The Inverse Of A Relation

Example Question #1 : Find The Inverse Of A Function

Example Question #2 : Find The Inverse Of A Function

Example Question #3 : Find The Inverse Of A Function

All Precalculus Resources

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