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Precalculus : Pre-Calculus

Study concepts, example questions & explanations for Precalculus

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12 Diagnostic Tests 380 Practice Tests Question of the Day Flashcards Learn by Concept

Example Questions

Example Question #2 : Find The Inverse Of A Relation

Find the inverse of the following function:

y=4x+10

Possible Answers:

$y=\frac{x-4}{10}$

$y=\frac{x-10}{4}$

y=4x-10

$y=\frac{x}{4}$

y=10x-4

Correct answer:

$y=\frac{x-10}{4}$

Explanation:

In order to find the inverse of the function, we need to switch the x- and y-variables.

y=4x+10

After switching the variables, we have the following:

x=4y+10

Now solve for the y-variable. Start by subtracting 10 from both sides of the equation.

x-10=4y+10-10

x-10=4y

Divide both sides of the equation by 4.

$\frac{x-10}{4}=\frac{4y}{4}$

Rearrange and solve.

$y=\frac{x-10}{4}$

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Example Question #1 : Find The Inverse Of A Function

Find the inverse of,

3x+1=y .

Possible Answers:

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

x=3y+1

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

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

Correct answer:

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

Explanation:

In order to find the inverse, switch the x and y variables in the function then solve for y.

3x+1=y

Switching variables we get,

3y+1=x .

Then solving for y to get our final answer.

3y=x-1

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

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Example Question #2 : Find The Inverse Of A Function

Find the inverse of,

$y=x^{2}$ .

Possible Answers:

$x=y^{2}$

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

$x=\pm \sqrt{y}$

$y=\pm\sqrt{x}$

Correct answer:

$y=\pm\sqrt{x}$

Explanation:

First, switch the variables making y=x^2 into $x=y^{2}$ .

Then solve for y by taking the square root of both sides.

y^2=x

$\sqrt {y^2}=\pm \sqrt x$

$y=\pm \sqrt x$

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Example Question #4 : Inverses

Find the inverse of the following equation.

$y=\sqrt{\ln(x)+2}$ .

Possible Answers:

$y=e^\pm ^{^{x^2}-2}$

$y=\pm x^2 -2$

y=x-2

y=e^x+2

$y=e^x^{^{2}-2}$

Correct answer:

$y=e^x^{^{2}-2}$

Explanation:

To find the inverse in this case, we need to switch our x and y variables and then solve for y.

Therefore,

$y=\sqrt{\ln(x)+2}$ becomes,

$x=\sqrt{\ln(y)+2}$

To solve for y we square both sides to get rid of the sqaure root.

$x^2=\ln(y)+2$

We then subtract 2 from both sides and take the exponenetial of each side, leaving us with the final answer.

$x^2-2=\ln(y)$

$e^{x^2-2}=e^{\ln(y)}$

$e^x^{^{2}-2}=y$

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Example Question #2 : Find The Inverse Of A Function

Find the inverse of the following function.

y=x^3+e^0

Possible Answers:

$y^{-1}=\sqrt[3]{x+e}$

$y^{-1}=\sqrt[{3}]{x-1}$

$y^{-1}=1-x^{3}}$

$y^{-1}=e^{0}}$

Correct answer:

$y^{-1}=\sqrt[{3}]{x-1}$

Explanation:

To find the inverse of y, or

$y^{-1}$

first switch your variables x and y in the equation.

$x_{i}=y_i^3+e^0$

Second, solve for the variable y_i in the resulting equation.

x_i-e^0=y_i^3

$(x_i-e^0)^\frac{1}{3}=(y_i^{3})^\frac{1}{3}$

$y_i=\sqrt[3]{x_i-e^0}$

Simplifying a number with 0 as the power, the inverse is

$y^{-1}=\sqrt[{3}]{x-1}$

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Example Question #181 : Algebra

Find the inverse of the following function.

$y=\ln\, x\: +\: \pi$

Possible Answers:

$y^{-1}=\pi-\ln(x)$

$y^{-1}=e^{x-\pi}$

$y^{-1}=\ln(y)-\pi$

Does not exist

Correct answer:

$y^{-1}=e^{x-\pi}$

Explanation:

To find the inverse of y, or

$y^{-1}$

first switch your variables x and y in the equation.

$x_i=\ln(y_i)+\pi$

Second, solve for the variable y_i in the resulting equation.

$x_i-\pi=\ln\,y_i$

And by setting each side of the equation as powers of base e,

$y^{-1}=e^{x-\pi}$

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Example Question #1 : Matrices

Find the inverse of the function.

y=x+5

Possible Answers:

$y^{-1}=-x-5$

$y^{-1}=x-5$

$y^{-1}=5+x$

$y^{-1}=5-x$

Correct answer:

$y^{-1}=x-5$

Explanation:

To find the inverse we need to switch the variables and then solve for y.

y=x+5

Switching the variables we get the following equation,

x=y+5 .

Now solve for y.

$y^{-1}=x-5$

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Example Question #4 : Find The Inverse Of A Function

Find the inverse of

y=10x+9

Possible Answers:

y=9x-10

y=10x

$y=\frac{x-9}{10}$

$y=\frac{9x}{10}$

y=9x+10

Correct answer:

$y=\frac{x-9}{10}$

Explanation:

So we first replace every with an and every with a .

Our resulting equation is:

x=10y+9

Now we simply solve for y.

Subtract 9 from both sides:

x-9=10y+9-9

x-9=10y

Now divide both sides by 10:

$\frac{x-9}{10}=\frac{10y}{10}$

$\frac{x-9}{10}=y$

The inverse of

y=10x+9

$y=\frac{x-9}{10}$

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Example Question #3 : Find The Inverse Of A Function

What is the inverse of

y=9x

Possible Answers:

x=0

y=0

y=x

$y=\frac{x}{9}$

y=9x

Correct answer:

$y=\frac{x}{9}$

Explanation:

To find the inverse of a function we just switch the places of all and with eachother.

y=9x

turns into

x=9y

Now we solve for

Divide both sides by

$\frac{x}{9}=\frac{9y}{9}$

$y=\frac{x}{9}$

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Example Question #3 : Find The Inverse Of A Function

If $f(x)=-\frac{3x}{5}+11$ , what is its inverse function, $f^{-1}(x)$ ?

Possible Answers:

$f^{-1}(x)=-\frac{3x}{5}+\frac{33}{5}$

$f^{-1}(x)=\frac{5x}{3}+\frac{55}{3}$

$f^{-1}(x)=-\frac{5x}{3}+\frac{55}{3}$

$f^{-1}(x)=-\frac{5x}{3}-\frac{55}{3}$

$f^{-1}(x)=-\frac{3x}{5}-\frac{33}{5}$

Correct answer:

$f^{-1}(x)=-\frac{5x}{3}+\frac{55}{3}$

Explanation:

We begin by taking $f(x)=-\frac{3x}{5}+11$ and changing the f(x) to a , giving us $y=-\frac{3x}{5}+11$ .

Next, we switch all of our x's and y's , giving us $x=-\frac{3y}{5}+11$ .

Finally, we solve for by subtracting from each side, multiplying each side by , and dividing each side by , leaving us with,

$y=-\frac{5x}{3}+\frac{55}{3}$ .

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Precalculus : Pre-Calculus

Study concepts, example questions & explanations for Precalculus

All Precalculus Resources

Example Questions

Example Question #2 : Find The Inverse Of A Relation

Example Question #1 : Find The Inverse Of A Function

Example Question #2 : Find The Inverse Of A Function

Example Question #4 : Inverses

Example Question #2 : Find The Inverse Of A Function

Example Question #181 : Algebra

Example Question #1 : Matrices

Example Question #4 : Find The Inverse Of A Function

Example Question #3 : Find The Inverse Of A Function

Example Question #3 : Find The Inverse Of A Function

All Precalculus Resources

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