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Algebra 3/4 : Algebra 3/4

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

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All Algebra 3/4 Resources

6 Practice Tests Question of the Day Flashcards Learn by Concept

Example Questions

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Example Question #1 : Functions & Relations

Find the composition, $f\circ g$ given

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

Possible Answers:

$f\circ g=9x^2+6x$

$f\circ g=3x^2+6$

$f\circ g=3x^2+6x$

$f\circ g=9x^2+6$

$f\circ g=3x^2+3x$

Correct answer:

$f\circ g=9x^2+6x$

Explanation:

To find the composition, $f\circ g$ given

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

recall what a composition of two functions represents.

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

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

Therefore, in this particular problem the composition becomes

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

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Example Question #2 : Functions & Relations

Find the inverse of f(x) .

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

Possible Answers:

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

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

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

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

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

Correct answer:

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

Explanation:

To find the inverse of a function swap the variables and solve for . 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.

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

$y=\frac{3}{2+5x}$

First, switch the variables.

$x=\frac{3}{2+5y}$

Now, perform algebraic operations to solve for .

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

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

$\\2+5y=\frac{3}{x} \\\\5y=\frac{3}{x}-2 \\\\y=\frac{3}{5x}-\frac{2}{5}$

Therefore, the inverse is

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

Report an Error

Example Question #3 : Functions & Relations

Find the composition, $f\circ g$ given

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

Possible Answers:

$f\circ g=14x^2+3$

$f\circ g=98x^2+3$

$f\circ g=49x^2+3x$

$f\circ g=32x^2+3$

$f\circ g=98x^2+3x$

Correct answer:

$f\circ g=98x^2+3$

Explanation:

To find the composition, $f\circ g$ given

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

recall what a composition of two functions represents.

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

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

Therefore, in this particular problem the composition becomes

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

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Example Question #4 : Functions & Relations

Find the inverse of the function f(x) .

f(x)=2x^2+4

Possible Answers:

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

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

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

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

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

Correct answer:

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

Explanation:

To find the inverse of a function swap the variables and solve for . 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.

f(x)=2x^2+4

y=2x^2+4

First, switch the variables.

x=2y^2+4

Now, perform algebraic operations to solve for .

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

Therefore, the inverse is

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

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Example Question #5 : Functions & Relations

Find the composition, $f\circ g$ given

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

Possible Answers:

$f\circ g=\frac{1}{2}x^2$

$f\circ g=\frac{1}{4}x^2$

$f\circ g=2x^2$

$f\circ g=\frac{1}{8}x^2$

$f\circ g=x^2$

Correct answer:

$f\circ g=\frac{1}{4}x^2$

Explanation:

To find the composition, $f\circ g$ given

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

recall what a composition of two functions represents.

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

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

Therefore, in this particular problem the composition becomes

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

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Example Question #6 : Functions & Relations

Find the composition, $f\circ g$ given

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

Possible Answers:

$f\circ g=9x^2+6x$

$f\circ g=9x^2+6$

$f\circ g=3x^2+6x$

$f\circ g=3x^2+6$

$f\circ g=3x^2+3x$

Correct answer:

$f\circ g=9x^2+6x$

Explanation:

To find the composition, $f\circ g$ given

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

recall what a composition of two functions represents.

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

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

Therefore, in this particular problem the composition becomes

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

Report an Error

Example Question #7 : Functions & Relations

Find the inverse of f(x) .

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

Possible Answers:

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

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

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

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

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

Correct answer:

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

Explanation:

To find the inverse of a function swap the variables and solve for . 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.

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

$y=\frac{3}{2+5x}$

First, switch the variables.

$x=\frac{3}{2+5y}$

Now, perform algebraic operations to solve for .

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

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

$\\2+5y=\frac{3}{x} \\\\5y=\frac{3}{x}-2 \\\\y=\frac{3}{5x}-\frac{2}{5}$

Therefore, the inverse is

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

Report an Error

Example Question #8 : Functions & Relations

Find the composition, $f\circ g$ given

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

Possible Answers:

$f\circ g=98x^2+3$

$f\circ g=49x^2+3x$

$f\circ g=98x^2+3x$

$f\circ g=14x^2+3$

$f\circ g=32x^2+3$

Correct answer:

$f\circ g=98x^2+3$

Explanation:

To find the composition, $f\circ g$ given

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

recall what a composition of two functions represents.

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

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

Therefore, in this particular problem the composition becomes

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

Report an Error

Example Question #9 : Functions & Relations

Find the inverse of the function f(x) .

f(x)=2x^2+4

Possible Answers:

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

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

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

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

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

Correct answer:

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

Explanation:

To find the inverse of a function swap the variables and solve for . 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.

f(x)=2x^2+4

y=2x^2+4

First, switch the variables.

x=2y^2+4

Now, perform algebraic operations to solve for .

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

Therefore, the inverse is

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

Report an Error

Example Question #1 : Simplify Radical Expressions

Simplify the following radical expression.

$\sqrt{125}+\sqrt{48}$

Possible Answers:

$5\sqrt{5}+2\sqrt{12}$

$25\sqrt{5}+16\sqrt{3}$

$5\sqrt{5}+4\sqrt{3}$

$25\sqrt{5}+4\sqrt{3}$

$5\sqrt{5}+8\sqrt{3}$

Correct answer:

$5\sqrt{5}+4\sqrt{3}$

Explanation:

To simplify the radical expression look at the factors under each radical.

$\sqrt{125}+\sqrt{48}$

$\sqrt{25\cdot 5}+\sqrt{4\cdot 12}$

Recall that 25 and 4 are perfect squares.

$5\sqrt{5}+2\sqrt{12}$

From here 12 can be factored further.

$\\5\sqrt{5}+2\sqrt{4\cdot 3} \\5\sqrt{5}+2\cdot 2\sqrt{3} \\5\sqrt{5}+4\sqrt{3}$

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Algebra 3/4 : Algebra 3/4

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

All Algebra 3/4 Resources

Example Questions

Example Question #1 : Functions & Relations

Example Question #2 : Functions & Relations

Example Question #3 : Functions & Relations

Example Question #4 : Functions & Relations

Example Question #5 : Functions & Relations

Example Question #6 : Functions & Relations

Example Question #7 : Functions & Relations

Example Question #8 : Functions & Relations

Example Question #9 : Functions & Relations

Example Question #1 : Simplify Radical Expressions

All Algebra 3/4 Resources

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