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All Common Core: High School - Functions Resources

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Flashcards: Inverse Functions Verified Through Composition: CCSS.Math.Content.HSF-BF.B.4b

Are f(x) and g(x) inverses of each other?

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

Yes

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Flashcards: Inverse Functions Verified Through Composition: CCSS.Math.Content.HSF-BF.B.4b

Are f(x) and g(x) inverses of each other?

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

Your answer:

Correct answer:

Yes

Explanation:

This question is testing one's ability to calculation of the composition of functions for the purpose of verifying inverse functions. It is important to recall that there are two compositions of functions that need to be calculated before two functions are verified as inverses.

For the purpose of Common Core Standards, verify by composition that one function is the inverse of another, 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: Calculate f(g(x)) .

Given

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

f(g(x)) can be found as follows.

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

First distribute the two in the numerator to the fraction.

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

Now multiply the one in the denominator by $\frac{2-x}{2-x}$ and add the two terms in the denominator together.

$f(g(x))=\frac{\left(\frac{2x}{2-x}\right)}{\left(\frac{x}{2-x} \right )+\frac{2-x}{2-x}}$

$f(g(x))=\frac{\left(\frac{2x}{2-x}\right)}{\left(\frac{x+2-x}{2-x} \right )}$

$f(g(x))=\frac{\left(\frac{2x}{2-x}\right)}{\left(\frac{2}{2-x} \right )}$

From here, multiple the numerator by the reciprocal of the denominator.

$f(g(x))=\left(\frac{2x}{2-x}\right)\times\left(\frac{2-x}{2} \right )$

The 2-x in the numerator and in the denominator cancel out as does the two.

$f(g(x))=\left(\frac{1x}{1}\right)\times\left(\frac{1}{1} \right )=x$

Step 2: Calculate g(f(x)) .

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

First multiply the two in the denominator by $\frac{x+1}{x+1}$ and then add the terms.

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

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

Now, multiply the numerator by the reciprocal of the denominator.

$g(f(x))=\left( \frac{2x}{x+1}\right)\times \left( \frac{x+1}{2}\right)$

The x+1 and the two cancel out.

$g(f(x))=\left( \frac{1x}{1}\right)\times \left( \frac{1}{1}\right)=x$

Step 3: Is f(g(x)) and g(f(x)) equal to ?

In order for two functions to be the inverse of one another the composition of their functions must equal . This is due to the fact that evaluating a function at the inverse function value are opposite operations and will cancel out to leave just .

Since both f(g(x)) and g(f(x)) equal to they are inverse functions of each other.

Flashcards: Inverse Functions Verified Through Composition: CCSS.Math.Content.HSF-BF.B.4b

Are f(x) and g(x) inverses of each other?

$\\ f(x)=x-2 \\g(x)=x+2$

Yes

Report an Error Copyright Notice

Flashcards: Inverse Functions Verified Through Composition: CCSS.Math.Content.HSF-BF.B.4b

Are f(x) and g(x) inverses of each other?

$\\ f(x)=x-2 \\g(x)=x+2$

Your answer:

Correct answer:

Yes

Explanation:

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: Calculate f(g(x)) .

Given

$\\ f(x)=x-2 \\g(x)=x+2$

f(g(x)) can be found as follows.

f(g(x))=(x+2)-2

First drop the parentheses.

f(g(x))=x+2-2

Now simplify by adding the constant terms together.

f(g(x))=x+2-2

f(g(x))=x

Step 2: Calculate g(f(x)) .

g(f(x))=(x-2)+2

First drop the parentheses.

g(f(x))=x-2+2

Now, add the constants together.

g(f(x))=x-2+2

g(f(x))=x

Step 3: Is f(g(x)) and g(f(x)) equal to ?

Since both f(g(x)) and g(f(x)) equal to they are inverse functions of each other.

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All Common Core: High School - Functions Resources

Flashcards: Inverse Functions Verified Through Composition: CCSS.Math.Content.HSF-BF.B.4b

Flashcards: Inverse Functions Verified Through Composition: CCSS.Math.Content.HSF-BF.B.4b

Flashcards: Inverse Functions Verified Through Composition: CCSS.Math.Content.HSF-BF.B.4b

Flashcards: Inverse Functions Verified Through Composition: CCSS.Math.Content.HSF-BF.B.4b

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