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Algebra II : Inverse Functions

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Example Question #91 : Inverse Functions

Choose the inverse of $y=\frac{x-3}{2}$ $\displaystyle y=\frac{x-3}{2}$ .

Possible Answers:

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

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

$y=2x+3$

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

y=3x-2 $\displaystyle y=3x-2$

Correct answer:

$y=2x+3$

Explanation:

To find the inverse of a linear function, switch the variables and solve for y.

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

Switch the variables:

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

Multiply both sides by 2:

2x=y-3 $\displaystyle 2x=y-3$

Add 3 to both sides to isolate y:

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

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Example Question #91 : Inverse Functions

Find the inverse function: $\displaystyle y=2(5-x)$

Possible Answers:

$y=\frac{1}{5}x-\frac{1}{10}$ $\displaystyle y=\frac{1}{5}x-\frac{1}{10}$

$y=\frac{1}{2}x-10$ $\displaystyle y=\frac{1}{2}x-10$

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

$y=\frac{1}{2}x-\frac{1}{10}$ $\displaystyle y=\frac{1}{2}x-\frac{1}{10}$

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

Correct answer:

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

Explanation:

Interchange the x and y-variables.

$\displaystyle x=2(5-y)$

Solve for y. Divide by two on both sides.

$\frac{x}{2}=\frac{2(5-y)}{2}$ $\displaystyle \frac{x}{2}=\frac{2(5-y)}{2}$

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

Add $\displaystyle y$ on both sides.

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

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

Subtract $\frac{x}{2}$ $\displaystyle \frac{x}{2}$ on both sides.

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

Simplify both sides.

The answer is: $y=5-\frac{1}{2}x$

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Algebra II : Inverse Functions

Study concepts, example questions & explanations for Algebra II

All Algebra II Resources

Example Questions

Example Question #91 : Inverse Functions

Example Question #91 : Inverse Functions

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