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Algebra II : Inverse Functions

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Example Questions

Example Question #31 : Inverse Functions

Find the inverse of the equation: y=6x-14

Possible Answers:

y=-6x-14

$y=\frac{1}{6}x -\frac{7}{3}$

$y=\frac{1}{6}x +\frac{7}{3}$

$y=-\frac{1}{6}x -\frac{7}{3}$

y=6x-14

Correct answer:

$y=\frac{1}{6}x +\frac{7}{3}$

Explanation:

To solve for the inverse, first interchange the x and y-variables.

x=6y-14

Solve for y. Add 14 on both sides.

x+14=6y-14+14

Simplify the right side.

x+14=6y

Divide by six on both sides.

$\frac{x+14}{6}=\frac{6y}{6}$

Simplify both sides of the equation.

The answer is: $y=\frac{1}{6}x +\frac{7}{3}$

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Example Question #32 : Inverse Functions

Find the inverse of the following function: $y=-\frac{2}{3}x+2$

Possible Answers:

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

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

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

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

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

Correct answer:

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

Explanation:

Interchange the x and y-variables.

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

Solve for y. Subtract two from both sides.

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

Simplify the right side.

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

Multiply both sides by the reciprocal of the coefficient in front of the y-variable.

$-\frac{3}{2}(x-2)=-\frac{2}{3}y\cdot -\frac{3}{2}$

Simplify both sides. Distribute the fraction on the left side.

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

The answer is: $y=-\frac{3}{2}x+3$

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Example Question #33 : Inverse Functions

Solve for the inverse: $y=\frac{1}{6}x+6$

Possible Answers:

y=6x-36

y=-6x-6

y=6x+36

y=-6x-36

y=-6x+6

Correct answer:

y=6x-36

Explanation:

Interchange the x and y-variables, and solve for y.

$x=\frac{1}{6}y+6$

Subtract six from both sides.

$x-6=\frac{1}{6}y+6-6$

Simplify the right side.

$x-6=\frac{1}{6}y$

In order to isolate the y-variable, we will need to multiply six on both sides.

$6(x-6)=\frac{1}{6}y\cdot 6$

Simplify both sides. Distribute the integer through the binomial on the left.

6x-36=y

The answer is: y=6x-36

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Example Question #31 : Inverse Functions

Find the inverse of the function: y=4x+3

Possible Answers:

$y=\frac{3}{4}x+\frac{3}{4}$

$y=\frac{3}{4}x-\frac{3}{4}$

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

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

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

Correct answer:

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

Explanation:

Interchange the x and y-variables.

x=4y+3

Solve for y. Subtract three from both sides.

x-3=4y+3-3

Simplify the right side.

x-3=4y

Divide by four on both sides.

$\frac{x-3}{4}=\frac{4y}{4}$

Simplify both sides.

The answer is: $y=\frac{1}{4}x-\frac{3}{4}$

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Example Question #35 : Inverse Functions

Find the inverse of: y=x^3+5

Possible Answers:

$y=\sqrt[3]{x}-5$

$y=-\sqrt[3]{x-5}$

$y=\sqrt[3]{3x}-5$

$y=\frac{\sqrt[3]{5}}{5}x+\frac{3}{5}$

$y=\sqrt[3]{x-5}$

Correct answer:

$y=\sqrt[3]{x-5}$

Explanation:

Interchange the x and y-variables. The equation becomes:

x=y^3+5

Subtract five from both sides.

x-5=y^3+5-5

x-5=y^3

Take the cubed root on both sides. This will eliminate the cubed exponent.

$\sqrt[3]{x-5}=\sqrt[3]{y^3}$

The answer is: $y=\sqrt[3]{x-5}$

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Example Question #36 : Inverse Functions

Determine the inverse of: $y=\frac{2}{5}x-\frac{1}{2}$

Possible Answers:

$y=\frac{5}{2}x+\frac{5}{4}$

$y=-\frac{5}{2}x-\frac{1}{2}$

$y=\frac{5}{2}x+\frac{1}{2}$

$y=\frac{5}{2}x-\frac{5}{4}$

$y=-\frac{5}{2}x+\frac{1}{2}$

Correct answer:

$y=\frac{5}{2}x+\frac{5}{4}$

Explanation:

Interchange the x and y-variables.

$x=\frac{2}{5}y-\frac{1}{2}$

Solve for y. Add one-half on both sides.

$x+\frac{1}{2}=\frac{2}{5}y-\frac{1}{2}+\frac{1}{2}$

Simplify both sides.

$x+\frac{1}{2}=\frac{2}{5}y$

Multiply five over two on both sides in order to isolate the y-variable.

$\frac{5}{2}(x+\frac{1}{2})=\frac{2}{5}y\cdot \frac{5}{2}$

Apply the distributive property on the left side. The right side will reduce to just a lone y-variable.

The answer is: $y=\frac{5}{2}x+\frac{5}{4}$

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Example Question #37 : Inverse Functions

Determine the inverse of: y=-2x-1

Possible Answers:

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

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

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

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

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

Correct answer:

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

Explanation:

Interchange the x and y-variables and solve for y.

x=-2y-1

Add one on both sides.

x+1=-2y-1+1

x+1=-2y

Divide by negative two on both sides.

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

Simplify the fractions.

The answer is: $y=-\frac{1}{2}x-\frac{1}{2}$

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Example Question #36 : Inverse Functions

Determine the inverse: y= 10x+3

Possible Answers:

$y=\frac{3}{10}x-3$

$y=\frac{1}{10}x-\frac{3}{10}$

$y=\frac{1}{10}x+\frac{3}{10}$

$y=\frac{1}{10}x-3$

$y=-\frac{1}{10}x-\frac{3}{10}$

Correct answer:

$y=\frac{1}{10}x-\frac{3}{10}$

Explanation:

In order to find the inverse of this function, interchange the x and y-variables.

x= 10y+3

Subtract three from both sides.

x-3= 10y+3-3

Simplify the equation.

x-3= 10y

Divide by ten on both sides.

$\frac{x-3}{10}= \frac{10y}{10}$

Simplify both sides.

The answer is: $y=\frac{1}{10}x-\frac{3}{10}$

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Example Question #38 : Inverse Functions

Determine the inverse for the function: y=-2x-30

Possible Answers:

$y=-\frac{1}{2}x+15$

$y=-\frac{1}{2}x-15$

$y=\frac{1}{2}x-30$

$y=-\frac{1}{2}x-30$

y=-2x+15

Correct answer:

$y=-\frac{1}{2}x-15$

Explanation:

To find the inverse function, swap the x and y-variables.

x=-2y-30

Solve for y. Add 30 on both sides.

x+30=-2y-30+30

Simplify the right side.

x+30=-2y

Divide by negative two on both sides.'

$\frac{x+30}{-2}=\frac{-2y}{-2}$

Simplify both fractions.

The answer is: $y=-\frac{1}{2}x-15$

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Example Question #231 : Functions And Graphs

Find the inverse function of the function below:

$f(x)=\frac{7x+3}{2x-1}$

Possible Answers:

$f^-^1(x)=\frac{2x-3}{7x+1}$

$f^-^1(x)=\frac{x+7}{2x-3}$

$f^-^1(x)=\frac{x+2}{7x-3}$

$f^-^1(x)=\frac{x+3}{2x-7}$

$f^-^1(x)=\frac{3x-7}{x+2}$

Correct answer:

$f^-^1(x)=\frac{x+3}{2x-7}$

Explanation:

To determine the inverse function f^-^1 of an explicitly defined function , substitute the dependent variable f(x)=y and independent variable for and respectively, and then solve the resultant equation for . This new equation will define the inverse function f^-^1 , provided that f(f^-^1(x))=x and f^-^1(f(x))=x for every in the domain of .

For this particular function, let denote the dependent variable f(x) :

$y=\frac{7x+3}{2x-1}$ .

Swap the variables and :

$x=\frac{7y+3}{2y-1}$ .

Let us now solve this equation for . Multiplying both sides by 2y-1 yields

2yx-x=7y+3 .

Subtracting the term and adding the term to both sides yields

2yx-7y=x+3 .

On the left-hand side of the equation, factor out from both terms using the distributive property to yield

y(2x-7)=x+3 .

Now divide both sides of the equation by 2x-7 to isolate the variable :

$y=\frac{x+3}{2x-7}$ .

In order to communicate the idea that this equation defines the inverse function to , let y=f^-^1(x) to yield the final answer:

$f^-^1(x)=\frac{x+3}{2x-7}$ .

To verify that this function is indeed the inverse of , calculate f(f^-^1(x)) and f^-^1(f(x)) .

$f(f^-^1(x))=\frac{7(\frac{x+3}{2x-7})+3}{2(\frac{x+3}{2x-7})-1}$

$=\frac{\frac{7x+21+6x-21}{2x-7}}{\frac{2x+6-2x+7}{2x-7}}$

$=\frac{13x}{13}$

$f^-^1(f(x))=\frac{(\frac{7x+3}{2x-1})+3}{2(\frac{7x+3}{2x-1})-7}$

$=\frac{\frac{7x+3+6x-3}{2x-1}}{\frac{14x+6-14x+7}{2x-1}}$

$=\frac{13x}{13}$

Hence, the inverse function of the function is

$f^-^1(x)=\frac{x+3}{2x-7}$

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Algebra II : Inverse Functions

Study concepts, example questions & explanations for Algebra II

All Algebra II Resources

Example Questions

Example Question #31 : Inverse Functions

Example Question #32 : Inverse Functions

Example Question #33 : Inverse Functions

Example Question #31 : Inverse Functions

Example Question #35 : Inverse Functions

Example Question #36 : Inverse Functions

Example Question #37 : Inverse Functions

Example Question #36 : Inverse Functions

Example Question #38 : Inverse Functions

Example Question #231 : Functions And Graphs

All Algebra II Resources

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