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Algebra II : Introduction to Functions

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Example Questions

Example Question #61 : Inverse Functions

Determine the inverse: y= 3(3-9x)

Possible Answers:

$y=-\frac{1}{27}x+\frac{1}{3}$

$y=-9x+\frac{1}{27}$

$y=-\frac{1}{3}x+\frac{1}{27}$

$y=-\frac{1}{27}x+\frac{1}{27}$

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

Correct answer:

$y=-\frac{1}{27}x+\frac{1}{3}$

Explanation:

Interchange the x and y-variables.

x= 3(3-9y)

Solve for y. Distribute the three through both terms of the binomial.

x=9-27y

Subtract 9 on both sides.

x-9=9-27y-9

The equation becomes: x-9=-27y

Divide by negative 27 on both sides.

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

Simplify both sides.

The answer is: $y=-\frac{1}{27}x+\frac{1}{3}$

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Example Question #261 : Introduction To Functions

Determine the inverse of: y=-4x-32

Possible Answers:

$y=-\frac{1}{4}x-8$

$y=-\frac{1}{4}x+\frac{1}{8}$

$y=-\frac{1}{4}x-32$

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

$y=-\frac{1}{4}x-\frac{1}{8}$

Correct answer:

$y=-\frac{1}{4}x-8$

Explanation:

Interchange the x and y-variables.

x=-4y-32

Solve for y. Add 32 on both sides.

x+32=-4y-32+32

The equation becomes:

x+32=-4y

Divide by negative four on both sides.

$\frac{x+32}{-4}=\frac{-4y}{-4}$

Simplify both sides.

The answer is: $y=-\frac{1}{4}x-8$

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Example Question #65 : Inverse Functions

Determine the inverse: y=2-(2x-3)

Possible Answers:

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

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

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

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

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

Correct answer:

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

Explanation:

Interchange the x and y variables.

x=2-(2y-3)

Solve for y.

Subtract two from both sides.

x-2=2-(2y-3)-2

x-2=-(2y-3)

Divide by negative one on both sides.

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

-x+2 =2y-3

Add three on both sides.

-x+2+3 =2y-3+3

-x+5 =2y

Divide by two on both sides.

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

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

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Example Question #66 : Inverse Functions

Determine the inverse of the function: y=-3(6x+3)

Possible Answers:

$y=-\frac{1}{9}x -\frac{1}{2}$

$y=-\frac{1}{18}x -\frac{1}{18}$

$y=-\frac{1}{9}x -\frac{1}{18}$

$y=-\frac{1}{18}x -\frac{1}{2}$

$y=-\frac{1}{9}x -\frac{1}{4}$

Correct answer:

$y=-\frac{1}{18}x -\frac{1}{2}$

Explanation:

Interchange the x and y variables.

x=-3(6y+3)

Solve for y. Distribute the negative three across both terms in the binomial.

x=-3(6y)+(-3)(3)

Simplify the equation.

x=-18y-9

Add 9 on both sides.

x+9=-18y-9+9

x+9=-18y

Divide by negative 18 on both sides.

$\frac{x+9}{-18}=\frac{-18y}{-18}$

The answer is: $y=-\frac{1}{18}x -\frac{1}{2}$

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Example Question #61 : Inverse Functions

Determine the inverse of the equation: y=-6x+3

Possible Answers:

$y=-\frac{1}{6}x+\frac{1}{4}$

$y=-\frac{1}{6}x+\frac{1}{2}$

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

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

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

Correct answer:

$y=-\frac{1}{6}x+\frac{1}{2}$

Explanation:

Interchange the x and y-variables.

x=-6y+3

Solve for y.

Subtract three from both sides.

x-3=-6y+3-3

x-3=-6y

Divide by negative 6 on both sides.

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

The answer is: $y=-\frac{1}{6}x+\frac{1}{2}$

Report an Error

Example Question #62 : Inverse Functions

Determine the inverse for: y=-6x-10

Possible Answers:

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

$y=-\frac{1}{6}x-\frac{10}{3}$

$y=-\frac{1}{3}x-\frac{5}{6}$

$y=-\frac{1}{6}x-\frac{5}{3}$

$y=-\frac{1}{3}x-\frac{5}{3}$

Correct answer:

$y=-\frac{1}{6}x-\frac{5}{3}$

Explanation:

Swap the x and y variables.

x=-6y-10

Solve for y. Add 10 on both sides.

x+10=-6y-10+10

x+10=-6y

Divide by negative six on both sides.

$\frac{x+10}{-6}=\frac{-6y}{-6}$

The answer is: $y=-\frac{1}{6}x-\frac{5}{3}$

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Example Question #69 : Inverse Functions

Determine the inverse: x+2y = -4

Possible Answers:

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

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

y=-2x -4

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

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

Correct answer:

y=-2x -4

Explanation:

In order to solve for the inverse, interchange the x and the y variables, and solve for y.

y+2x = -4

Subtract from both sides of the equation.

y+2x-2x =-2x -4

The answer is: y=-2x -4

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Example Question #63 : Inverse Functions

Relation

Above is the graph of a function f(x) . Which choice gives the graph of $f^{-1}(x)$ ?

Possible Answers:

Relation

Correct answer:

Relation

Explanation:

Given the graph of f(x) , the graph of its inverse, $f^{-1}(x)$ is the reflection of the former about the line y = x . This line is in dark green below; critical points are reflected as shown:

Relation

The blue figure is $f^{-1}(x)$ , recreated below:

Relation

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Example Question #71 : Inverse Functions

Define a function $f(x) = \frac{1}{5}x +7$ .

Which statement correctly gives $f^{-1}(x)$ ?

Possible Answers:

$f^{-1}(x) = \frac{5}{7}x-7$

$f^{-1}(x) = \frac{5}{7}x-35$

None of the other choices gives the correct response.

$f^{-1}(x) =5x-7$

$f^{-1}(x) =5x-35$

Correct answer:

$f^{-1}(x) =5x-35$

Explanation:

The inverse function $f^{-1}(x)$ of a function f(x) can be found as follows:

Replace f(x) with :

$f(x) = \frac{1}{5}x +7$

$y= \frac{1}{5}x +7$

Switch the positions of and :

$x= \frac{1}{5}y +7$

or,

$\frac{1}{5}y +7 = x$

Solve for - that is, isolate it on one side.

Subtract 7:

$\frac{1}{5}y +7 - 7 = x - 7$

$\frac{1}{5}y = x - 7$

Multiply by 5, distributing on the right:

$5 \cdot \frac{1}{5}y =5 \cdot (x - 7)$

$y =5 \cdot x - 5 \cdot 7$

y =5x-35

Replace with $f^{-1}(x)$ :

$f^{-1}(x) =5x-35$

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Example Question #72 : Inverse Functions

Define a function $f(x) = 5^{x} + 5$ .

Which statement correctly gives $f^{-1}(x)$ ?

Possible Answers:

$f^{-1}(x) = \log_{5} x - 1$

$f^{-1}(x) = \log_{5} x - 5$

$f^{-1}(x) = \log_{5} (x - 5)$

$f^{-1}(x) =-5 \log_{5}x$

$f^{-1}(x) = \frac{1}{5} \log_{5}x$

Correct answer:

$f^{-1}(x) = \log_{5} (x - 5)$

Explanation:

The inverse function $f^{-1}(x)$ of a function f(x) can be found as follows:

Replace f(x) with :

$f(x) = 5^{x} + 5$

$y = 5^{x} + 5$

Switch the positions of and :

$x = 5^{y} + 5$ ,

$5^{y} + 5 = x$

Solve for - that is, isolate it on one side - as follows:

First, subtract 5 from both sides:

$5^{y} + 5- 5 = x - 5$

$5^{y} = x - 5$

Take the base-5 logarithm of both sides:

$\log_{5} 5^{y} = \log_{5} (x - 5)$

A property of logarithms states that $\log_{a} a^{N}= N$ , so

$y = \log_{5} (x - 5)$

Replace with $f^{-1}(x)$ :

$f^{-1}(x) = \log_{5} (x - 5)$

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Algebra II : Introduction to Functions

Study concepts, example questions & explanations for Algebra II

All Algebra II Resources

Example Questions

Example Question #61 : Inverse Functions

Example Question #261 : Introduction To Functions

Example Question #65 : Inverse Functions

Example Question #66 : Inverse Functions

Example Question #61 : Inverse Functions

Example Question #62 : Inverse Functions

Example Question #69 : Inverse Functions

Example Question #63 : Inverse Functions

Example Question #71 : Inverse Functions

Example Question #72 : Inverse Functions

All Algebra II Resources

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