Algebra 3/4 : Functions & Relations

Study concepts, example questions & explanations for Algebra 3/4

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Example Questions

Example Question #1 : Functions & Relations

Find the composition, \(\displaystyle f\circ g\) given

\(\displaystyle \\f(x)=x^2+2x \\g(x)=3x\)

Possible Answers:

\(\displaystyle f\circ g=3x^2+6x\)

\(\displaystyle f\circ g=9x^2+6\)

\(\displaystyle f\circ g=3x^2+6\)

\(\displaystyle f\circ g=3x^2+3x\)

\(\displaystyle f\circ g=9x^2+6x\)

Correct answer:

\(\displaystyle f\circ g=9x^2+6x\)

Explanation:

To find the composition, \(\displaystyle f\circ g\) given

\(\displaystyle \\f(x)=x^2+2x \\g(x)=3x\)

recall what a composition of two functions represents.

\(\displaystyle f \circ g=f(g(x))\)

This means that the function \(\displaystyle g(x)\) will replace each \(\displaystyle x\) in the function \(\displaystyle f(x)\).

Therefore, in this particular problem the composition becomes

\(\displaystyle \\f\circ g=f(g(x)) \\f(g(x))=(3x)^2+2(3x) \\f(g(x))=9x^2+6x\)

Example Question #2 : Functions & Relations

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

\(\displaystyle f(x)=\frac{3}{2+5x}\)

Possible Answers:

\(\displaystyle f^{-1}(x)=\frac{15}{x}-\frac{2}{5}\)

\(\displaystyle f^{-1}(x)=\frac{3}{5x}-\frac{2}{5}\)

\(\displaystyle f^{-1}(x)=\frac{15}{x}+\frac{2}{5}\)

\(\displaystyle f^{-1}(x)=\frac{3}{5x}+\frac{2}{5}\)

\(\displaystyle f^{-1}(x)=\frac{3}{5x}-10\)

Correct answer:

\(\displaystyle f^{-1}(x)=\frac{3}{5x}-\frac{2}{5}\)

Explanation:

To find the inverse of a function swap the variables and solve for \(\displaystyle y\). The function and its inverse when multiplied together, equals one. This means that the inverse undoes the function. 

For this particular function the inverse is found as follows.

\(\displaystyle f(x)=\frac{3}{2+5x}\)

\(\displaystyle y=\frac{3}{2+5x}\)

First, switch the variables.

\(\displaystyle x=\frac{3}{2+5y}\)

Now, perform algebraic operations to solve for \(\displaystyle y\).

\(\displaystyle x\cdot (2+5y)={3}\)

\(\displaystyle \frac{x\cdot(2+5y)}{x}=\frac{3}{x}\)

\(\displaystyle \\2+5y=\frac{3}{x} \\\\5y=\frac{3}{x}-2 \\\\y=\frac{3}{5x}-\frac{2}{5}\)

Therefore, the inverse is

\(\displaystyle f^{-1}(x)=\frac{3}{5x}-\frac{2}{5}\)

Example Question #3 : Functions & Relations

Find the composition, \(\displaystyle f\circ g\) given

\(\displaystyle \\f(x)=2x^2+3 \\g(x)=7x\)

Possible Answers:

\(\displaystyle f\circ g=32x^2+3\)

\(\displaystyle f\circ g=98x^2+3x\)

\(\displaystyle f\circ g=49x^2+3x\)

\(\displaystyle f\circ g=98x^2+3\)

\(\displaystyle f\circ g=14x^2+3\)

Correct answer:

\(\displaystyle f\circ g=98x^2+3\)

Explanation:

To find the composition, \(\displaystyle f\circ g\) given

\(\displaystyle \\f(x)=2x^2+3 \\g(x)=7x\)

recall what a composition of two functions represents.

\(\displaystyle f \circ g=f(g(x))\)

This means that the function \(\displaystyle g(x)\) will replace each \(\displaystyle x\) in the function \(\displaystyle f(x)\).

Therefore, in this particular problem the composition becomes

\(\displaystyle \\f\circ g=f(g(x)) \\f(g(x))=2(7x)^2+3 \\f(g(x))=2(49x^2)+3\\f(g(x))=98x^2+3\)

Example Question #4 : Algebra 3/4

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

\(\displaystyle f(x)=2x^2+4\)

Possible Answers:

\(\displaystyle f^{-1}(x)=\sqrt{2x-8}\)

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

\(\displaystyle f^{-1}(x)=\sqrt{\frac{x}{2}+4}\)

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

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

Correct answer:

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

Explanation:

To find the inverse of a function swap the variables and solve for \(\displaystyle y\). The function and its inverse when multiplied together, equals one. This means that the inverse undoes the function. 

For this particular function the inverse is found as follows.

\(\displaystyle f(x)=2x^2+4\)

\(\displaystyle y=2x^2+4\)

First, switch the variables.

\(\displaystyle x=2y^2+4\)

Now, perform algebraic operations to solve for \(\displaystyle y\).

\(\displaystyle \\x-4=2y^2 \\\\\frac{x-4}{2}=y^2 \\\\\frac{x}{2}-2=y^2 \\\\\sqrt{\frac{x}{2}-2}=y\)

Therefore, the inverse is

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

Example Question #4 : Functions & Relations

Find the composition, \(\displaystyle f\circ g\) given

\(\displaystyle \\f(x)=x^2 \\\\g(x)=\frac{1}{2}x\)

Possible Answers:

\(\displaystyle f\circ g=2x^2\)

\(\displaystyle f\circ g=x^2\)

\(\displaystyle f\circ g=\frac{1}{2}x^2\)

\(\displaystyle f\circ g=\frac{1}{8}x^2\)

\(\displaystyle f\circ g=\frac{1}{4}x^2\)

Correct answer:

\(\displaystyle f\circ g=\frac{1}{4}x^2\)

Explanation:

To find the composition, \(\displaystyle f\circ g\) given

\(\displaystyle \\f(x)=x^2 \\\\g(x)=\frac{1}{2}x\)

recall what a composition of two functions represents.

\(\displaystyle f \circ g=f(g(x))\)

This means that the function \(\displaystyle g(x)\) will replace each \(\displaystyle x\) in the function \(\displaystyle f(x)\).

Therefore, in this particular problem the composition becomes

\(\displaystyle \\f\circ g=f(g(x)) \\\\f(g(x))=\left(\frac{1}{2}x \right )^2\\\\f(g(x))=\frac{1}{4}x^2\)

Example Question #5 : Functions & Relations

Find the composition, \(\displaystyle f\circ g\) given

\(\displaystyle \\f(x)=x^2+2x \\g(x)=3x\)

Possible Answers:

\(\displaystyle f\circ g=9x^2+6\)

\(\displaystyle f\circ g=3x^2+3x\)

\(\displaystyle f\circ g=3x^2+6x\)

\(\displaystyle f\circ g=9x^2+6x\)

\(\displaystyle f\circ g=3x^2+6\)

Correct answer:

\(\displaystyle f\circ g=9x^2+6x\)

Explanation:

To find the composition, \(\displaystyle f\circ g\) given

\(\displaystyle \\f(x)=x^2+2x \\g(x)=3x\)

recall what a composition of two functions represents.

\(\displaystyle f \circ g=f(g(x))\)

This means that the function \(\displaystyle g(x)\) will replace each \(\displaystyle x\) in the function \(\displaystyle f(x)\).

Therefore, in this particular problem the composition becomes

\(\displaystyle \\f\circ g=f(g(x)) \\f(g(x))=(3x)^2+2(3x) \\f(g(x))=9x^2+6x\)

Example Question #6 : Functions & Relations

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

\(\displaystyle f(x)=\frac{3}{2+5x}\)

Possible Answers:

\(\displaystyle f^{-1}(x)=\frac{15}{x}-\frac{2}{5}\)

\(\displaystyle f^{-1}(x)=\frac{15}{x}+\frac{2}{5}\)

\(\displaystyle f^{-1}(x)=\frac{3}{5x}-10\)

\(\displaystyle f^{-1}(x)=\frac{3}{5x}+\frac{2}{5}\)

\(\displaystyle f^{-1}(x)=\frac{3}{5x}-\frac{2}{5}\)

Correct answer:

\(\displaystyle f^{-1}(x)=\frac{3}{5x}-\frac{2}{5}\)

Explanation:

To find the inverse of a function swap the variables and solve for \(\displaystyle y\). The function and its inverse when multiplied together, equals one. This means that the inverse undoes the function. 

For this particular function the inverse is found as follows.

\(\displaystyle f(x)=\frac{3}{2+5x}\)

\(\displaystyle y=\frac{3}{2+5x}\)

First, switch the variables.

\(\displaystyle x=\frac{3}{2+5y}\)

Now, perform algebraic operations to solve for \(\displaystyle y\).

\(\displaystyle x\cdot (2+5y)={3}\)

\(\displaystyle \frac{x\cdot(2+5y)}{x}=\frac{3}{x}\)

\(\displaystyle \\2+5y=\frac{3}{x} \\\\5y=\frac{3}{x}-2 \\\\y=\frac{3}{5x}-\frac{2}{5}\)

Therefore, the inverse is

\(\displaystyle f^{-1}(x)=\frac{3}{5x}-\frac{2}{5}\)

Example Question #7 : Functions & Relations

Find the composition, \(\displaystyle f\circ g\) given

\(\displaystyle \\f(x)=2x^2+3 \\g(x)=7x\)

Possible Answers:

\(\displaystyle f\circ g=14x^2+3\)

\(\displaystyle f\circ g=98x^2+3\)

\(\displaystyle f\circ g=32x^2+3\)

\(\displaystyle f\circ g=98x^2+3x\)

\(\displaystyle f\circ g=49x^2+3x\)

Correct answer:

\(\displaystyle f\circ g=98x^2+3\)

Explanation:

To find the composition, \(\displaystyle f\circ g\) given

\(\displaystyle \\f(x)=2x^2+3 \\g(x)=7x\)

recall what a composition of two functions represents.

\(\displaystyle f \circ g=f(g(x))\)

This means that the function \(\displaystyle g(x)\) will replace each \(\displaystyle x\) in the function \(\displaystyle f(x)\).

Therefore, in this particular problem the composition becomes

\(\displaystyle \\f\circ g=f(g(x)) \\f(g(x))=2(7x)^2+3 \\f(g(x))=2(49x^2)+3\\f(g(x))=98x^2+3\)

Example Question #8 : Functions & Relations

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

\(\displaystyle f(x)=2x^2+4\)

Possible Answers:

\(\displaystyle f^{-1}(x)=\sqrt{2x-8}\)

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

\(\displaystyle f^{-1}(x)=\sqrt{\frac{x}{2}+4}\)

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

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

Correct answer:

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

Explanation:

To find the inverse of a function swap the variables and solve for \(\displaystyle y\). The function and its inverse when multiplied together, equals one. This means that the inverse undoes the function. 

For this particular function the inverse is found as follows.

\(\displaystyle f(x)=2x^2+4\)

\(\displaystyle y=2x^2+4\)

First, switch the variables.

\(\displaystyle x=2y^2+4\)

Now, perform algebraic operations to solve for \(\displaystyle y\).

\(\displaystyle \\x-4=2y^2 \\\\\frac{x-4}{2}=y^2 \\\\\frac{x}{2}-2=y^2 \\\\\sqrt{\frac{x}{2}-2}=y\)

Therefore, the inverse is

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

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