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Common Core: High School - Functions : Simple Functions and Coresponding Inverses: CCSS.Math.Content.HSF-BF.B.4a

Study concepts, example questions & explanations for Common Core: High School - Functions

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All Common Core: High School - Functions Resources

6 Diagnostic Tests 82 Practice Tests Question of the Day Flashcards Learn by Concept

Example Questions

Example Question #61 : Building Functions

Find the inverse of f(x) .

f(x)=3x-7

Possible Answers:

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

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

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

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

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

Correct answer:

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

Explanation:

This question is testing one's ability to algebraically calculate the inverse of a given function. This also builds one's understanding of the concept of a function and its inverse at a basic level, graphing functions, and the Cartesian plane and its coordinate system.

For the purpose of Common Core Standards, finding the inverse of a simple function, falls within the Cluster B of build new functions from existing functions concept (CCSS.Math.content.HSF.BF.B).

Knowing the standard and the concept for which it relates to, we can now do the step-by-step process to solve the problem in question.

Step 1: Switch the and variables.

The given function is,

f(x)=3x-7

recall that f(x)=y therefore,

y=3x-7 .

Now switch the variables.

x=3y-7

Step 2: Solve for .

Solving for requires the use of algebraic operations to move constants from side to side. Remember to use the opposite operation to move a constant from one side to the other.

$\\x=3y-7 \\x{\color{Blue} +7}=3y-7{\color{Blue} +7} \\x+7=3y \\ \\ \frac{x+7}{3}=\frac{3y}{3} \\ \\ \frac{x+7}{3}=y\Rightarrow y=\frac{1}{3}x+\frac{7}{3}$

Step 3: Answer the question.

Recall that after the variable are switch, and is solved for it is really the inverse of that is being solved for thus, $y=f^{-1}(x)$ .

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

Report an Error

Example Question #1 : Simple Functions And Coresponding Inverses: Ccss.Math.Content.Hsf Bf.B.4a

Find the inverse of f(x) .

f(x)=2x+5

Possible Answers:

$f^{-1}(x)=\frac{1}{2}x-\frac{5}{2}$

$f^{-1}(x)=-\frac{1}{2}x-\frac{5}{2}$

$f^{-1}(x)=\frac{1}{2}x-\frac{5}{4}$

$f^{-1}(x)=\frac{1}{2}x+\frac{5}{2}$

$f^{-1}(x)=\frac{1}{4}x-\frac{5}{4}$

Correct answer:

$f^{-1}(x)=\frac{1}{2}x-\frac{5}{2}$

Explanation:

For the purpose of Common Core Standards, finding the inverse of a simple function, falls within the Cluster B of build new functions from existing functions concept (CCSS.Math.content.HSF.BF.B).

Knowing the standard and the concept for which it relates to, we can now do the step-by-step process to solve the problem in question.

Step 1: Switch the and variables.

The given function is,

f(x)=2x+5

recall that f(x)=y therefore,

y=2x+5 .

Now switch the variables.

x=2y+5

Step 2: Solve for .

Solving for requires the use of algebraic operations to move constants from side to side. Remember to use the opposite operation to move a constant from one side to the other.

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

Step 3: Answer the question.

Recall that after the variable are switch, and is solved for it is really the inverse of that is being solved for thus, $y=f^{-1}(x)$ .

$f^{-1}(x)=\frac{1}{2}x-\frac{5}{2}$

Report an Error

Example Question #2 : Simple Functions And Coresponding Inverses: Ccss.Math.Content.Hsf Bf.B.4a

Find the inverse of f(x) .

f(x)=x+10

Possible Answers:

$f^{-1}(x)=10x+1$

$f^{-1}(x)=10x-1$

$f^{-1}(x)=x+10$

$f^{-1}(x)=\frac{x}{10}$

$f^{-1}(x)=x-10$

Correct answer:

$f^{-1}(x)=x-10$

Explanation:

For the purpose of Common Core Standards, finding the inverse of a simple function, falls within the Cluster B of build new functions from existing functions concept (CCSS.Math.content.HSF.BF.B).

Knowing the standard and the concept for which it relates to, we can now do the step-by-step process to solve the problem in question.

Step 1: Switch the and variables.

The given function is,

f(x)=x+10

recall that f(x)=y therefore,

y=x+10 .

Now switch the variables.

x=y+10

Step 2: Solve for .

Solving for requires the use of algebraic operations to move constants from side to side. Remember to use the opposite operation to move a constant from one side to the other.

$\\x=y+10 \\x{\color{Blue} -10}=y+10{\color{Blue} -10} \\x-10=y$

Step 3: Answer the question.

Recall that after the variable are switch, and is solved for it is really the inverse of that is being solved for thus, $y=f^{-1}(x)$ .

$f^{-1}(x)=x-10$

Report an Error

Example Question #1 : Simple Functions And Coresponding Inverses: Ccss.Math.Content.Hsf Bf.B.4a

Find the inverse of f(x) .

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

Possible Answers:

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

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

$f^{-1}(x)=-2x-1$

$f^{-1}(x)=2x-1$

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

Correct answer:

$f^{-1}(x)=2x-1$

Explanation:

For the purpose of Common Core Standards, finding the inverse of a simple function, falls within the Cluster B of build new functions from existing functions concept (CCSS.Math.content.HSF.BF.B).

Knowing the standard and the concept for which it relates to, we can now do the step-by-step process to solve the problem in question.

Step 1: Switch the and variables.

The given function is,

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

recall that f(x)=y therefore,

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

Now switch the variables.

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

Step 2: Solve for .

Solving for requires the use of algebraic operations to move constants from side to side. Remember to use the opposite operation to move a constant from one side to the other.

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

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

${\color{Blue} -1}+2x=y+1{\color{Blue} -1}$

2x-1=y

Step 3: Answer the question.

Recall that after the variable are switch, and is solved for it is really the inverse of that is being solved for thus, $y=f^{-1}(x)$ .

$f^{-1}(x)=2x-1$

Report an Error

Example Question #2 : Simple Functions And Coresponding Inverses: Ccss.Math.Content.Hsf Bf.B.4a

Find the inverse of f(x) .

$f(x)=\frac{x}{2}+10$

Possible Answers:

$f^{-1}(x)=2x+20$

$f^{-1}(x)=2x+10$

$f^{-1}(x)=2x-10$

$f^{-1}(x)=2x-20$

$f^{-1}(x)=-2x-10$

Correct answer:

$f^{-1}(x)=2x-20$

Explanation:

For the purpose of Common Core Standards, finding the inverse of a simple function, falls within the Cluster B of build new functions from existing functions concept (CCSS.Math.content.HSF.BF.B).

Knowing the standard and the concept for which it relates to, we can now do the step-by-step process to solve the problem in question.

Step 1: Switch the and variables.

The given function is,

$f(x)=\frac{x}{2}+10$

recall that f(x)=y therefore,

$y=\frac{x}{2}+10$ .

Now switch the variables.

$x=\frac{y}{2}+10$

Step 2: Solve for .

Solving for requires the use of algebraic operations to move constants from side to side. Remember to use the opposite operation to move a constant from one side to the other.

$x=\frac{y}{2}+10$

${\color{Blue} -10}+x=\frac{y}{2}+10{\color{Blue} -10}$

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

$2(x-10)=\frac{y}{2}\cdot 2$

2x-20={y}

Step 3: Answer the question.

Recall that after the variable are switch, and is solved for it is really the inverse of that is being solved for thus, $y=f^{-1}(x)$ .

$f^{-1}(x)=2x-20$

Report an Error

Example Question #5 : Simple Functions And Coresponding Inverses: Ccss.Math.Content.Hsf Bf.B.4a

Find the inverse of f(x) .

f(x)=x+1

Possible Answers:

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

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

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

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

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

Correct answer:

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

Explanation:

For the purpose of Common Core Standards, finding the inverse of a simple function, falls within the Cluster B of build new functions from existing functions concept (CCSS.Math.content.HSF.BF.B).

Knowing the standard and the concept for which it relates to, we can now do the step-by-step process to solve the problem in question.

Step 1: Switch the and variables.

The given function is,

f(x)=x+1

recall that f(x)=y therefore,

y=x+1 .

Now switch the variables.

x=y+1

Step 2: Solve for .

Solving for requires the use of algebraic operations to move constants from side to side. Remember to use the opposite operation to move a constant from one side to the other.

$\\{\color{Blue} -1}+x=y+1{\color{Blue} -1} \\x-1=y$

Step 3: Answer the question.

Recall that after the variable are switch, and is solved for it is really the inverse of that is being solved for thus, $y=f^{-1}(x)$ .

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

Report an Error

Example Question #2 : Simple Functions And Coresponding Inverses: Ccss.Math.Content.Hsf Bf.B.4a

Find the inverse of f(x) .

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

Possible Answers:

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

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

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

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

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

Correct answer:

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

Explanation:

For the purpose of Common Core Standards, finding the inverse of a simple function, falls within the Cluster B of build new functions from existing functions concept (CCSS.Math.content.HSF.BF.B).

Knowing the standard and the concept for which it relates to, we can now do the step-by-step process to solve the problem in question.

Step 1: Switch the and variables.

The given function is,

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

recall that f(x)=y therefore,

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

Now switch the variables.

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

Step 2: Solve for .

Solving for requires the use of algebraic operations to move constants from side to side. Remember to use the opposite operation to move a constant from one side to the other.

$\\{\color{Blue} 3\cdot}x=\frac{1}{3}y{\color{Blue} \cdot 3} \\3x=y$

Step 3: Answer the question.

Recall that after the variable are switch, and is solved for it is really the inverse of that is being solved for thus, $y=f^{-1}(x)$ .

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

Report an Error

Example Question #7 : Simple Functions And Coresponding Inverses: Ccss.Math.Content.Hsf Bf.B.4a

Find the inverse of f(x) .

$f(x)=\frac{1}{3}x-6$

Possible Answers:

$f^{-1}(x)=3x+18$

$f^{-1}(x)=-3x-6$

$f^{-1}(x)=-3x+18$

$f^{-1}(x)=-3x+6$

$f^{-1}(x)=3x-18$

Correct answer:

$f^{-1}(x)=3x+18$

Explanation:

For the purpose of Common Core Standards, finding the inverse of a simple function, falls within the Cluster B of build new functions from existing functions concept (CCSS.Math.content.HSF.BF.B).

Knowing the standard and the concept for which it relates to, we can now do the step-by-step process to solve the problem in question.

Step 1: Switch the and variables.

The given function is,

$f(x)=\frac{1}{3}x-6$

recall that f(x)=y therefore,

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

Now switch the variables.

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

Step 2: Solve for .

Solving for requires the use of algebraic operations to move constants from side to side. Remember to use the opposite operation to move a constant from one side to the other.

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

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

${\color{Blue} 3}(x+6)=\frac{1}{3}y{\color{Blue} \cdot 3}$

3x+18=y

Step 3: Answer the question.

Recall that after the variable are switch, and is solved for it is really the inverse of that is being solved for thus, $y=f^{-1}(x)$ .

$f^{-1}(x)=3x+18$

Report an Error

Example Question #1 : Simple Functions And Coresponding Inverses: Ccss.Math.Content.Hsf Bf.B.4a

Find the inverse of f(x) .

f(x)=3x

Possible Answers:

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

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

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

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

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

Correct answer:

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

Explanation:

For the purpose of Common Core Standards, finding the inverse of a simple function, falls within the Cluster B of build new functions from existing functions concept (CCSS.Math.content.HSF.BF.B).

Knowing the standard and the concept for which it relates to, we can now do the step-by-step process to solve the problem in question.

Step 1: Switch the and variables.

The given function is,

f(x)=3x

recall that f(x)=y therefore,

y=3x .

Now switch the variables.

x=3y

Step 2: Solve for .

Solving for requires the use of algebraic operations to move constants from side to side. Remember to use the opposite operation to move a constant from one side to the other.

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

Step 3: Answer the question.

Recall that after the variable are switch, and is solved for it is really the inverse of that is being solved for thus, $y=f^{-1}(x)$ .

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

Report an Error

Example Question #9 : Simple Functions And Coresponding Inverses: Ccss.Math.Content.Hsf Bf.B.4a

Find the inverse of f(x) .

f(x)=-x+2

Possible Answers:

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

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

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

$f^{-1}(x)=-x-2$

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

Correct answer:

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

Explanation:

For the purpose of Common Core Standards, finding the inverse of a simple function, falls within the Cluster B of build new functions from existing functions concept (CCSS.Math.content.HSF.BF.B).

Knowing the standard and the concept for which it relates to, we can now do the step-by-step process to solve the problem in question.

Step 1: Switch the and variables.

The given function is,

f(x)=-x+2

recall that f(x)=y therefore,

y=-x+2 .

Now switch the variables.

x=-y+2

Step 2: Solve for .

Solving for requires the use of algebraic operations to move constants from side to side. Remember to use the opposite operation to move a constant from one side to the other.

$\\x=-y+2 \\x{\color{Blue} -2}=-y+2{\color{Blue} -2} \\{\color{Blue} -1}(x-2)=-y{\color{Blue} (-1)} \\-x+2=y$

Step 3: Answer the question.

Recall that after the variable are switch, and is solved for it is really the inverse of that is being solved for thus, $y=f^{-1}(x)$ .

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

Report an Error

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Common Core: High School - Functions : Simple Functions and Coresponding Inverses: CCSS.Math.Content.HSF-BF.B.4a

Study concepts, example questions & explanations for Common Core: High School - Functions

All Common Core: High School - Functions Resources

Example Questions

Example Question #61 : Building Functions

Example Question #1 : Simple Functions And Coresponding Inverses: Ccss.Math.Content.Hsf Bf.B.4a

Example Question #2 : Simple Functions And Coresponding Inverses: Ccss.Math.Content.Hsf Bf.B.4a

Example Question #1 : Simple Functions And Coresponding Inverses: Ccss.Math.Content.Hsf Bf.B.4a

Example Question #2 : Simple Functions And Coresponding Inverses: Ccss.Math.Content.Hsf Bf.B.4a

Example Question #5 : Simple Functions And Coresponding Inverses: Ccss.Math.Content.Hsf Bf.B.4a

Example Question #2 : Simple Functions And Coresponding Inverses: Ccss.Math.Content.Hsf Bf.B.4a

Example Question #7 : Simple Functions And Coresponding Inverses: Ccss.Math.Content.Hsf Bf.B.4a

Example Question #1 : Simple Functions And Coresponding Inverses: Ccss.Math.Content.Hsf Bf.B.4a

Example Question #9 : Simple Functions And Coresponding Inverses: Ccss.Math.Content.Hsf Bf.B.4a

All Common Core: High School - Functions Resources

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