Algebra II - Inverse Functions

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

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

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

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

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

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

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

Determine the inverse of:

$y=\frac{1}{7}x+\frac{4}{7}$

$y=\frac{1}{7}x-\frac{4}{7}$

$y=-7x+\frac{4}{7}$

$y=\frac{4}{7}x-\frac{4}{7}$

Explanation

Interchange the x and y variables and solve for y.

Add four on both sides.

Simplify both sides.

Divide by seven on both sides.

$\frac{x+4}{7}= \frac{7y}{7}$

The answer is: $y=\frac{1}{7}x+\frac{4}{7}$

Untitled

The above table shows a function with domain $\left { -3, -2, -1, 0, 1, 2, 3 \right }$ .

True or false: has an inverse function.

False

True

Explanation

A function has an inverse function if and only if, for all in the domain of , if , it follows that . In other words, no two values in the domain can be matched with the same range value.

If we order the rows by range value, we see this to not be the case:

Untitled

and . Since two range values exist to which more than one domain value is matched, the function has no inverse.

Determine the inverse:

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

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

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

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

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

Explanation

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

Subtract three from both sides.

Simplify the equation.

Divide by ten on both sides.

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

Simplify both sides.

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

Determine the inverse function given the equation:

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

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

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

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

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

Explanation

Interchange the x and y variables.

Solve for y. Add six on both sides.

Divide both sides by negative four.

$\frac{x+6}{-4}= \frac{-4y}{-4}$

The answer is: $y=-\frac{1}{4}x-\frac{3}{2}$

Determine the inverse of:

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

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

$y=\frac{1}{9}x-\frac{1}{9}$

$y=-\frac{1}{9}x-\frac{1}{9}$

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

Explanation

Switch the x and y variables.

Solve for y. Subtract 27 on both sides.

Divide both sides by 9.

$\frac{x-27}{9}=\frac{9y}{9}$

The answer is: $y=\frac{1}{9}x-3$

Determine the inverse of:

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

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

$y=\frac{1}{3}x+\frac{1}{12}$

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

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

Explanation

Interchange the x and y variables.

Solve for y.

Add 12 on both sides of the equation.

Divide both sides by three.

$\frac{x+12}{3}=\frac{3y}{3}$

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

Find the inverse function:

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

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

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

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

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

Explanation

Interchange the x and y-variables.

Solve for y. Divide by two on both sides.

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

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

Add on both sides.

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

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

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

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

Simplify both sides.

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

Determine the inverse function:

$y=\frac{1}{9}x-\frac{26}{9}$

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

$y=-9x-\frac{26}{9}$

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

$y=-\frac{1}{9}x+\frac{26}{9}$

Explanation

Interchange the x and y-variables.

Solve for y. Subtract 26 from both sides.

The equation becomes:

Divide by nine on both sides.

$\frac{x-26}{9}=\frac{9y}{9}$

Simplify both fractions.

The answer is: $y=\frac{1}{9}x-\frac{26}{9}$

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

Which of the following represents $f^{-1}(x)$ ?

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

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

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

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

Explanation

The question is asking for the inverse function. To find the inverse, first switch input and output -- which is usually easiest if you use notation instead of . Then, solve for .