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Algebra II : Introduction to Functions

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Example Questions

Example Question #23 : Inverse Functions

What is the inverse function of $f(x)=\frac{x^2 + 4x + 3}{x + 1}$ ?

Possible Answers:

$\frac{x+1}{x^2 + 4x + 3}$

x-3

$\frac{1}{x-1}$

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

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

Correct answer:

x-3

Explanation:

Before we begin, it would help to try and simplify the problem. Let's start by factoring the numerator:

x^2 + 4x + 3 = (x+1)(x+3)

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

We can see that the (x+1) terms will cancel, making this problem much easier:

f(x)= x+3

To find the inverse, we can put a where the is currently, put a for the f(x) , and then solve for .

x = y + 3

y = x-3

$f(x)^{-1}=x-3$

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Example Question #741 : Algebra Ii

Find the inverse of $f(x)= \frac{4}{x^2 +3}$ .

Possible Answers:

$3-\sqrt{\frac{4}{x}}$

$\pm\sqrt{\frac{4}{x}-3}$

4(x+2)^2

$\frac{x^{2}+3}{4}$

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

Correct answer:

$\pm\sqrt{\frac{4}{x}-3}$

Explanation:

First we're going to change the in the function to a , set the function equal to , and solve for :

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

Multiply each side by y^2 + 3 :

x(y^2 + 3)=4

Divide each side by :

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

Subtract from each side:

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

Finally, take the square root of each side:

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

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Example Question #742 : Algebra Ii

Find the inverse of the given function:

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

Possible Answers:

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

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

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

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

Correct answer:

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

Explanation:

To find the inverse of the function, we must first replace all x in the equation with y, and all y in the equation with x:

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

Now, solve for y:

x(y+1)=y+3

xy+x=y+3

xy-y=3-x

y(x-1)=3-x

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

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Example Question #26 : Inverse Functions

Find the inverse function of f(x)=x^3-1 .

Possible Answers:

$y=\sqrt[3]{x+1}$

$y=\sqrt[3]{-x+1}$

$y=\sqrt[3]{x-1}$

There is no inverse for f(x)

Correct answer:

$y=\sqrt[3]{x+1}$

Explanation:

Step 1: To find the inverse of a function f(x) , we will first switch the places of x and y. Wherever we saw y, we put x. Where we see x, we put y.

y=x^3+1

After changing places of letters:

x=y^3+1

Step 2: We want to solve for y, so we ned to get rid of the on the right side.

x-1=y^3

Step 3: We want to find , but we have y^3 . To find , we will take the cube root of both sides

$\sqrt[3]{x-1}=\sqrt[3]{y^3}$

This simplifies to:

$\sqrt[3]{x-1}=y$

The inverse function of f(x) is $\sqrt[3]{x-1}=y$

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Example Question #221 : Introduction To Functions

What is the inverse for the function $y=\frac{x-7}{8}$

Possible Answers:

$y=\frac{8}{x-7}$

y=8x-7

y=x+7

Inverse does not exist

y=8x+7

Correct answer:

y=8x+7

Explanation:

To find the inverse of a function switch the position of x and y and solve to y.

$x=\frac{y-7}{8}\Rightarrow8x=y-7\Rightarrow8x+7=y$

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Example Question #28 : Inverse Functions

Find the inverse of: y=4(x-6)

Possible Answers:

y=-4x+24

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

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

y=-4x-24

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

Correct answer:

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

Explanation:

Interchange the x and y-variables.

x=4(y-6)

Solve for y. Distribute the constant through the binomial.

x=4y-24

Add 24 on both sides.

x+24=4y-24+24

The equation becomes:

x+24=4y

Divide by four on both sides.

$\frac{x+24}{4}=\frac{4y}{4}$

Simplify both sides.

The answer is: $y=\frac{1}{4}x+6$

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Example Question #29 : Inverse Functions

Find $f^{-1}(x)$ if f(x)= 9x-3 .

Possible Answers:

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

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

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

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

$y=\sqrt{9x-3}$

Correct answer:

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

Explanation:

To solve for the inverse function, first replace f(x) with .

The equation becomes: y= 9x-3

Interchange the variables.

x=9y-3

Add three on both sides.

x+3=9y

Divide by nine on both sides.

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

Simplify both sides of the equation.

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

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Example Question #22 : Inverse Functions

Find the inverse of the function: $y=-\frac{1}{4}x-6$

Possible Answers:

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

y=-4x-24

y=-4x+24

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

$\textup{The inverse is not given.}$

Correct answer:

y=-4x-24

Explanation:

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

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

Solve for y. Add 6 on both sides of the equation.

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

The equation becomes:

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

Multiply by negative four on both sides to isolate the y-variable.

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

Simplify both sides of the equation.

The answer is: y=-4x-24

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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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Algebra II : Introduction to Functions

Study concepts, example questions & explanations for Algebra II

All Algebra II Resources

Example Questions

Example Question #23 : Inverse Functions

Example Question #741 : Algebra Ii

Example Question #742 : Algebra Ii

Example Question #26 : Inverse Functions

Example Question #221 : Introduction To Functions

Example Question #28 : Inverse Functions

Example Question #29 : Inverse Functions

Example Question #22 : Inverse Functions

Example Question #31 : Inverse Functions

Example Question #32 : Inverse Functions

All Algebra II Resources

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