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Precalculus : Pre-Calculus

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Example Questions

Example Question #3 : Determine If A Relation Is A Function

Is the following relation of ordered pairs a function?

(1,1); (2,2); (3,1); (4,2); (5,1); (6,7)

Possible Answers:

Cannot be determined

Yes

Correct answer:

Yes

Explanation:

A set of ordered pairs is a function if it passes the vertical line test.

Because there are no more than one corresponding value for any given value, the relation of ordered pairs IS a function.

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Example Question #1 : Define A Field

Which of the following is not true about a field. (Note: the real numbers $\small \mathbb{R}$ is a field)

Possible Answers:

There is an element $\small e$ in the field such that $\small e+a=a+e=a$ for any element $\small a$ in the field.

For every element $\small a$ in the field, there is another element $\small b$ such that their product $\small \small ab$ is equal to $\small 1$ , where $\small 1$ is the multiplicative identity, $\small 1$ in the case of real numbers.

A field can be defined in many ways.

For every element $\small a$ in the field, there is another element $\small b$ such that their sum $\small a+b$ is equal to $\small 0$ , where $\small 0$ is the additive identity.

We have $\small ab=ba$ for any $\small a$ and $\small b$ in the field.

Correct answer:

Explanation:

It is not the case that for any element $\small a$ in a field, there is another one $\small b$ such that their product is $\small 1$ . Take $\small 0$ in the real numbers. Multiply $\small 0$ by any number and you get $\small 0$ , so you will never get $\small 1$ . This is true for any field that has more than 1 element.

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Example Question #1 : Modeling

John lives in Atlanta, but commutes every Monday to LaGrange where he has an apartment he stays in Monday-Friday for work. Each Monday he drives 350 miles to LaGrange. Once he arrives to his home away from home he is in walking distance of work and does not use his car for anything else. After 23 weeks his odometer shows 186,000 miles. Write an equation that models his odometer reading as a function of the number of weeks he has been driving after commencing his new job.

Possible Answers:

y=350x-169,900

y=350x+169,900

y=700x-169,900

y=700x+186,000

y=700x+169,900

Correct answer:

y=700x+169,900

Explanation:

The rate of change of his mileage is 700 per week (350 x 2=700 there and back). The rate of change is the same thing as slope. Since we are looking for equation an equation that models his odometer reading as a function of the number of weeks he has been driving we can extract the point (23 , 18600) since after 23 weeks his odometer read 18,600 miles. Now we will use the point slope formula:

y-186000=700(x-23)

distribute the right side

y-186000=700x-16100

isolate y

y=700x+169,900

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Example Question #1 : Modeling

John lives in Atlanta, but commutes every Monday to LaGrange where he has an apartment he stays in Monday-Friday. Each Monday he drives 350 miles to LaGrange. Once he arrives to his apartment he is in walking distance of work and does not use his car for anything else. After 23 weeks his odometer shows 186,000 miles. Write an equation that models his odometer reading as a function of the number of weeks he has been driving after commencing his new job. Using the equation you just made, what is the y intercept or his original mileage before starting?

Possible Answers:

Need more information to solve.

y intercept=169,900 miles

y intercept=186,000 miles

y intercept=153,800 miles

y intercept=16,100 miles

Correct answer:

y intercept=169,900 miles

Explanation:

The y intercept can be found by plugging in 0 for x in your original equation: y=700x+169,900 because at x=0 you can only reach the y axis and thus will find the y intercept.

y=700(0)+169900

y=169,900 miles

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Example Question #1 : Inverse Functions

What is the inverse function of

y=3x+5 ?

Possible Answers:

y=x-5

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

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

y=3x+5

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

Correct answer:

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

Explanation:

To find the inverse function of

y=3x+5

we replace the with and vice versa.

x=3y+5

Now solve for

x-5=3y

3y=x-5

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

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

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Example Question #2 : Find The Inverse Of A Relation

Find the inverse of the function.

$f(x)= \sin{\frac{x^2}{10}}$

Possible Answers:

$f^{-1}(x)=\sqrt{ 10 \arcsin{x} }$

$f^{-1}(x)=10 \sqrt{ \arcsin{y} }$

$f^{-1}(x)=\arcsin{ \frac{x^2}{10} }$

$f^{-1}(x)=\sqrt{ 10 \cos{x} }$

Correct answer:

$f^{-1}(x)=\sqrt{ 10 \arcsin{x} }$

Explanation:

To find the inverse function, first replace f(x) with :

$y= \sin{ \frac{x^2}{10} }$

Now replace each with an and each with a :

$x= \sin{ \frac{y^2}{10} }$

Solve the above equation for :

$\arcsin{x}= \frac{y^2}{10}$

$\sqrt{10 \arcsin{x}} = y$

Replace with $f^{-1}(x)$ . This is the inverse function:

$f^{-1}(x)=\sqrt{ 10 \arcsin{x} }$

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Example Question #2 : Linear Algebra

Find the inverse of the function.

$g(x)= \frac{x+3}{2x-4}$

Possible Answers:

$g^{-1}(x)= \frac{4x+3}{2x-1}$

$g^{-1}(x)= \frac{3x+4}{x-2}$

$g^{-1}(x)= \frac{x+3}{-4x+1}$

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

Correct answer:

$g^{-1}(x)= \frac{4x+3}{2x-1}$

Explanation:

To find the inverse function, first replace g(x) with :

$y= \frac{x+3}{2x-4}$

Now replace each with an and each with a :

$x= \frac{y+3}{2y-4}$

Solve the above equation for :

x(2y-4)=y+3

2xy-4x=y+3

2xy-y=4x+3

(2x-1)y=4x+3

$y= \frac{4x+3}{2x-1}$

Replace with $g^{-1}(x)$ . This is the inverse function:

$g^{-1}(x)= \frac{4x+3}{2x-1}$

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Example Question #3 : Find The Inverse Of A Relation

Find the inverse of the function y=2x^2+1 .

Possible Answers:

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

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

f^-^1(x)=x+1

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

f^-^1(x)=-x+1

Correct answer:

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

Explanation:

To find the inverse of y=2x^2+1 , interchange the and terms and solve for .

y=2x^2+1

x=2y^2+1

x-1=2y^2

$\frac{x-1}{2}=y^2$

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

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Example Question #1 : Find The Inverse Of A Relation

What point is the inverse of the (2,3) ?

Possible Answers:

(2,-3)

(-2,-3)

(-2,3)

(3,2)

Correct answer:

(3,2)

Explanation:

When trying to find the inverse of a point, switch the x and y values.

So (2,3) $\rightarrow (3,2)$

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Example Question #2 : Find The Inverse Of A Relation

What is the inverse of $\left(\frac{1}{2},3\right)$ ?

Possible Answers:

$\left(3,-\frac{1}{2}\right)$

$\left(-3,-\frac{1}{2}\right)$

$\left(-3,\frac{1}{2}\right)$

$\left(3,\frac{1}{2}\right)$

Correct answer:

$\left(3,\frac{1}{2}\right)$

Explanation:

When trying to find the inverse of a point, switch the x and y values.

So, $\left(\frac{1}{2},3\right)\rightarrow \left(3, \frac{1}{2} \right )$

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Precalculus : Pre-Calculus

Study concepts, example questions & explanations for Precalculus

All Precalculus Resources

Example Questions

Example Question #3 : Determine If A Relation Is A Function

Example Question #1 : Define A Field

Example Question #1 : Modeling

Example Question #1 : Modeling

Example Question #1 : Inverse Functions

Example Question #2 : Find The Inverse Of A Relation

Example Question #2 : Linear Algebra

Example Question #3 : Find The Inverse Of A Relation

Example Question #1 : Find The Inverse Of A Relation

Example Question #2 : Find The Inverse Of A Relation

All Precalculus Resources

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