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Flashcards: Simple Functions and Coresponding Inverses: CCSS.Math.Content.HSF-BF.B.4a

Find the inverse of f(x) $\displaystyle f(x)$ .

$\displaystyle f(x)=3x-7$

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

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

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

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

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

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All Common Core: High School - Functions Resources

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Flashcards: Simple Functions and Coresponding Inverses: CCSS.Math.Content.HSF-BF.B.4a

Find the inverse of f(x) $\displaystyle f(x)$ .

$\displaystyle f(x)=3x-7$

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

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

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

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

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

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Flashcards: Simple Functions and Coresponding Inverses: CCSS.Math.Content.HSF-BF.B.4a

Find the inverse of f(x) $\displaystyle f(x)$ .

$\displaystyle f(x)=3x-7$

Your answer:

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

Your answer:

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

Your answer:

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

Your answer:

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

Correct answer:

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

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 $\displaystyle x$ and $\displaystyle y$ variables.

The given function is,

$\displaystyle f(x)=3x-7$

recall that f(x)=y $\displaystyle f(x)=y$ therefore,

y=3x-7 $\displaystyle y=3x-7$ .

Now switch the variables.

x=3y-7 $\displaystyle x=3y-7$

Step 2: Solve for $\displaystyle y$ .

Solving for $\displaystyle y$ 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.

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

Step 3: Answer the question.

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

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

Flashcards: Simple Functions and Coresponding Inverses: CCSS.Math.Content.HSF-BF.B.4a

Find the inverse of f(x) $\displaystyle f(x)$ .

$\displaystyle f(x)=2x+5$

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

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

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

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

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

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Flashcards: Simple Functions and Coresponding Inverses: CCSS.Math.Content.HSF-BF.B.4a

Find the inverse of f(x) $\displaystyle f(x)$ .

$\displaystyle f(x)=2x+5$

Your answer:

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

Your answer:

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

Your answer:

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

Your answer:

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

Correct answer:

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

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 $\displaystyle x$ and $\displaystyle y$ variables.

The given function is,

$\displaystyle f(x)=2x+5$

recall that f(x)=y $\displaystyle f(x)=y$ therefore,

y=2x+5 $\displaystyle y=2x+5$ .

Now switch the variables.

x=2y+5 $\displaystyle x=2y+5$

Step 2: Solve for $\displaystyle y$ .

Solving for $\displaystyle y$ 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.

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

Step 3: Answer the question.

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

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

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Flashcards: Simple Functions and Coresponding Inverses: CCSS.Math.Content.HSF-BF.B.4a

All Common Core: High School - Functions Resources

Flashcards: Simple Functions and Coresponding Inverses: CCSS.Math.Content.HSF-BF.B.4a

Flashcards: Simple Functions and Coresponding Inverses: CCSS.Math.Content.HSF-BF.B.4a

Flashcards: Simple Functions and Coresponding Inverses: CCSS.Math.Content.HSF-BF.B.4a

Flashcards: Simple Functions and Coresponding Inverses: CCSS.Math.Content.HSF-BF.B.4a

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