Call Now to Set Up Tutoring:

(888) 888-0446

Calculus 1 : Calculus

Study concepts, example questions & explanations for Calculus 1

Create An Account

All Calculus 1 Resources

10 Diagnostic Tests 438 Practice Tests Question of the Day Flashcards Learn by Concept

Example Questions

Example Question #51 : How To Find The Meaning Of Functions

Find dv/dt if:

$v=\sin(2x);x=2t^2-5t+9$

Possible Answers:

$(4t-5)\cos(4t^2+5t-9)$

$(8t+10)\cos(4t^2-10t-18)$

$(8t-10)\cos(4t^2-10t+18)$

$(4t+5)\sin(2t^2-5t+9)$

$(4t^2-10t+18)\cos(2t)$

Correct answer:

$(8t-10)\cos(4t^2-10t+18)$

Explanation:

Solving for dv/dt, requires use of the chain rule:

$\frac{dv}{dt}=\frac{dv}{dx}\frac{dx}{dt}=\frac{d}{dx}[\sin(2x)]\frac{d}{dt}[2t^2-5t+9]$

$\frac{dv}{dt}=(2\cos(2x))(4t-5)=(4t-5)(2\cos(2(2t^2-5t+9)))$

$\frac{dv}{dt}=(8t-10)\cos(4t^2-10t+18)$

This is one of the answer choices.

Report an Error

Example Question #51 : How To Find The Meaning Of Functions

Find

$\lim_{x\rightarrow 1} \frac{6x^2-7x+1}{x^3-1}$ .

Possible Answers:

$-\infty$

$+\infty$

$\frac{5}{3}$

Undefined

Correct answer:

$\frac{5}{3}$

Explanation:

If you plug into the equation, you get $\frac{0}{0}$ , meaning you should use L'Hopital's Rule to find the limit.

L'Hopital's Rule states that if

$\lim_{x \to c}f(x)=\lim_{x \to c}g(x)=0 \text{ or } \pm\infty$ and that if

$\lim_{x\to c}\frac{f'(x)}{g'(x)}$ exists,

then

$\lim_{x\to c}\frac{f(x)}{g(x)} = \lim_{x\to c}\frac{f'(x)}{g'(x)}$

In this case:

f(x) = 6x^2 -7x+1

g(x) = x^3-1

Start by finding the derivative of the numerator and evaluating it at :

$\frac{\mathrm{d} }{\mathrm{d} x}6x^2-7x + 1=12x-7=5$

Do the same for the denominator:

$\frac{\mathrm{d} }{\mathrm{d} x}x^3-1 = 3x^2 = 3$

The limit is then equal to the derivative of the numerator evaluated at divided by the derivative of the denominator evaluated at , or $\frac{5}{3}$ .

Report an Error

Example Question #61 : How To Find The Meaning Of Functions

Find the slope of the tangent line equation for $y = \sin (2x) \cdot x^2$ when $x = \frac{\pi}{2}$ .

Possible Answers:

$\pi$

$- \frac{\pi^2}{4}$

$- \frac{\pi^2}{2}$

$-2 \pi$

Correct answer:

$- \frac{\pi^2}{2}$

Explanation:

To find the slope, we need to differentiate the given function. By product rule and chain rule, we have $y' = 2 \cos (2x) \cdot x^2 + \sin (2x) \cdot (2x).$ $\Rightarrow y'(\frac{\pi}{2}) = 2 \cos {\pi} \cdot \frac{\pi^2}{4} + \sin \pi \cdot \pi \\ = 2 \cdot (-1) \cdot \frac{\pi^2}{4} + 0 \cdot \pi = - \frac{\pi^2}{2} .$

Report an Error

Example Question #61 : How To Find The Meaning Of Functions

Function

What is a function?

Possible Answers:

A function is a matematical equation with one or, more variables

A function is a relationship that assigns to each input value a single output value

A function is a relationship that can produce multiple output values for each single input value

None of the above

A function is an equation with at least two unknowns such as x and y

Correct answer:

A function is a relationship that assigns to each input value a single output value

Explanation:

You could think of a function as a machine that takes in some number, performs an operation on it, and then spits out another number

Report an Error

Example Question #62 : How To Find The Meaning Of Functions

Identify Function

Which one of the following is not a function?

Possible Answers:

$f(x)=\sqrt{x}(\sqrt{x}-x\sqrt{x})$

$19x^{2}+157x-\frac{5}{2}=y$

$y^{2}=x+1$

$f(x)=5x^{5}+2$

$ax^{2}+bx+c=y$

Correct answer:

$y^{2}=x+1$

Explanation:

For any value of in $y^{2}=x+1$ , there will be two values of . So it is not a function.

Report an Error

Example Question #63 : How To Find The Meaning Of Functions

Type of Functions

What type of function is this:

$f(x)=\begin{Bmatrix} x^2+1 & if \, \, x\leqslant 3\\ 12-x& if \, \, x>3\end{Bmatrix}$

Possible Answers:

Linear

Quadratic

Polynomial

Piecewise

Rational

Correct answer:

Piecewise

Explanation:

Piecewise functions are a special type function in which the formula changes for different values.

Report an Error

Example Question #65 : How To Find The Meaning Of Functions

Find the domain of the function $f(x) = \frac {\ln x}{\sqrt {x^2-1}}.$

Possible Answers:

x>1

-1 < x <1

x>1 or x < -1

0 < x < 1

x > 0 or x < -1

Correct answer:

x>1

Explanation:

$\ln(x)$ is defined when x>0 , and $\sqrt {x^2-1}$ is defined when $x^2-1 \geq 0$ . Since the radical part is in the denominator, it cannot be . Therefore, we need $x^2 -1 >0 \Rightarrow x>1 \or \ x<-1.$ Combining this domain with x>0 , we get x>1 .

Report an Error

Example Question #66 : How To Find The Meaning Of Functions

$f(x)= 3x^{3}+7x-2$

What is the slope of this curve at x=4 ?

Possible Answers:

152

150

151

220

218

150

152

151

218

220

Correct answer:

151

Explanation:

To find the slope of a curve at a certain point, you must first find the derivative of that curve.

The derivative of this curve is $f'(x)=9x^{2}-7$ .

Then, plug in for into the derivative and you will get 151 as the slope of f(x) at x=2 .

Report an Error

Example Question #66 : How To Find The Meaning Of Functions

What are the critical points of f(x)=x^3-2x^2+x ?

Possible Answers:

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

x= 1

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

$x=\frac{1}{3}, 1$

Correct answer:

$x=\frac{1}{3}, 1$

Explanation:

To find the critical points of a function, you need to first find the derivative and then set that equal to 0. To find the derivative, multiply the exponent by the leading coefficient and then subtract 1 from the exponent. Therefore, the derivative is $f{}'(x)=3x^2-4x+1$ . Set that equal to 0 and then factor so that you get: (3x-1)(x-1)=0 . Solve for x in both expressions so that your answer is: $x=\frac{1}{3}, 1$ .

Report an Error

Example Question #64 : How To Find The Meaning Of Functions

Finding when the second derivative is positive tells us what about a function?

Possible Answers:

When the function is increasing.

When the function is concave up.

When the function is concave down.

When the function is decreasing.

Correct answer:

When the function is concave up.

Explanation:

The first derivative of a function tells us about the slope and the second derivative tells us about concavity. Thus, when the second derivative is positive, it tell us that the concavity is upward. Therefore, a postive second derivative means the function is concave up.

Report an Error

View Calculus Tutors

Kurosh
Certified Tutor

University of tehran, Bachelor of Engineering, Electrical Engineering.

View Calculus Tutors

Chelsea
Certified Tutor

The University of Texas at Dallas, Bachelor of Science, Physics. The University of Texas at Dallas, Master of Science, Bioinf...

View Calculus Tutors

Shane
Certified Tutor

The University of Texas at Dallas, Bachelor of Science, Computer Engineering, General.

All Calculus 1 Resources

10 Diagnostic Tests 438 Practice Tests Question of the Day Flashcards Learn by Concept

Popular Subjects

GRE Tutors in Dallas Fort Worth, Computer Science Tutors in Washington DC, MCAT Tutors in Denver, Physics Tutors in San Diego, ISEE Tutors in Denver, Math Tutors in Dallas Fort Worth, MCAT Tutors in New York City, ISEE Tutors in Atlanta, Statistics Tutors in San Diego, ISEE Tutors in Dallas Fort Worth

Popular Courses & Classes

SAT Courses & Classes in Atlanta, SSAT Courses & Classes in Philadelphia, GMAT Courses & Classes in Washington DC, GRE Courses & Classes in Miami, Spanish Courses & Classes in Washington DC, Spanish Courses & Classes in Phoenix, Spanish Courses & Classes in Dallas Fort Worth, ISEE Courses & Classes in Atlanta, ACT Courses & Classes in San Diego, LSAT Courses & Classes in Boston

Popular Test Prep

SAT Test Prep in Washington DC, SSAT Test Prep in Los Angeles, SAT Test Prep in Philadelphia, SSAT Test Prep in Chicago, LSAT Test Prep in Chicago, ISEE Test Prep in Phoenix, MCAT Test Prep in Dallas Fort Worth, SSAT Test Prep in Seattle, GRE Test Prep in Seattle, SSAT Test Prep in San Diego

DMCA Complaint

If you believe that content available by means of the Website (as defined in our Terms of Service) infringes one or more of your copyrights, please notify us by providing a written notice (“Infringement Notice”) containing the information described below to the designated agent listed below. If Varsity Tutors takes action in response to an Infringement Notice, it will make a good faith attempt to contact the party that made such content available by means of the most recent email address, if any, provided by such party to Varsity Tutors.

Your Infringement Notice may be forwarded to the party that made the content available or to third parties such as ChillingEffects.org.

Please be advised that you will be liable for damages (including costs and attorneys’ fees) if you materially misrepresent that a product or activity is infringing your copyrights. Thus, if you are not sure content located on or linked-to by the Website infringes your copyright, you should consider first contacting an attorney.

Please follow these steps to file a notice:

You must include the following:

A physical or electronic signature of the copyright owner or a person authorized to act on their behalf; An identification of the copyright claimed to have been infringed; A description of the nature and exact location of the content that you claim to infringe your copyright, in \ sufficient detail to permit Varsity Tutors to find and positively identify that content; for example we require a link to the specific question (not just the name of the question) that contains the content and a description of which specific portion of the question – an image, a link, the text, etc – your complaint refers to; Your name, address, telephone number and email address; and A statement by you: (a) that you believe in good faith that the use of the content that you claim to infringe your copyright is not authorized by law, or by the copyright owner or such owner’s agent; (b) that all of the information contained in your Infringement Notice is accurate, and (c) under penalty of perjury, that you are either the copyright owner or a person authorized to act on their behalf.

Send your complaint to our designated agent at:

Charles Cohn Varsity Tutors LLC
101 S. Hanley Rd, Suite 300
St. Louis, MO 63105

Or fill out the form below:

Do not fill in this field

Contact Information

* Full legal name

Company name

* Phone number

* Email address

Address

* Address1

Address2

* City

* State

* Zipcode

* Country

Complaint Details

* URL of copyrighted work (where your original material is located)

* Please describe the copyrighted work so that it may be easily identified

I am the owner, or an agent authorized to act on behalf of the owner of an exclusive right that is allegedly infringed.

I have a good faith belief that the use of the material in the manner complained of is not authorized by the copyright owner, its agent, or the law.

This notification is accurate.

I acknowledge that there may be adverse legal consequences for making false or bad faith allegations of copyright infringement by using this process.

* Signature (your digital signature is legally binding)

HIGH SCHOOL

GRADUATE SCHOOL

K-8

Search 50+ Tests

math tutoring

science tutoring

foreign languages

elementary tutoring

other

Search 350+ Subjects

Calculus 1 : Calculus

Study concepts, example questions & explanations for Calculus 1

All Calculus 1 Resources

Example Questions

Example Question #51 : How To Find The Meaning Of Functions

Find dv/dt if:

Example Question #51 : How To Find The Meaning Of Functions

Example Question #61 : How To Find The Meaning Of Functions

Example Question #61 : How To Find The Meaning Of Functions

Example Question #62 : How To Find The Meaning Of Functions

Example Question #63 : How To Find The Meaning Of Functions

Example Question #65 : How To Find The Meaning Of Functions

Example Question #66 : How To Find The Meaning Of Functions

Example Question #66 : How To Find The Meaning Of Functions

Example Question #64 : How To Find The Meaning Of Functions

All Calculus 1 Resources

Report an issue with this question

DMCA Complaint

Contact Information

Address

Complaint Details

Email address:
Your name:
Feedback: