We are open Saturday and Sunday!

Call Now to Set Up Tutoring:

(888) 888-0446

Calculus AB : Analytical Applications of Derivatives

Study concepts, example questions & explanations for Calculus AB

Create An Account

All Calculus AB Resources

45 Practice Tests Question of the Day Flashcards Learn by Concept

Example Questions

← Previous 1 2 3 4 5 6 7 8 Next →

Example Question #2 : Use Mean Value Theorem, Extreme Value Theorem, And Critical Points

Find the area under the curve $f(x)=\frac{1}{\sqrt{x+2}}$ between $2\leq x\leq 7$ .

Possible Answers:

$3$

$2$

$4$

$5$

$1$

Correct answer:

$2$

Explanation:

To find the area under the curve, we need to integrate. In this case, it is a definite integral.

$\int_{2}^{7}\frac{1}{\sqrt{x+2}}dx=2\sqrt{x+2}\Big|_2^7=2$

Report an Error

Example Question #3 : Use Mean Value Theorem, Extreme Value Theorem, And Critical Points

Find the area bounded by $y=2, y=x, y=\frac{1}{9}x^2, x=3$

Possible Answers:

2.5

1.5

3.5

Correct answer:

Explanation:

The easiest way to look at this is to plot the graphs. The shaded area is the actual area that we want to compute. We can first find area bounded by x=3 and y=2 in the first quadrant and subtract the excessive areas. The area of that rectangle box is 6. The area under the curve $\frac{1}{9}x^{2}$ is $\int_{0}^{3}\frac{1}{9}x^2=1$ .

The area of the triangle above the curve y=x is 2. Therefore, the area bounded is 6-1-2=3 .

Screen shot 2020 09 04 at 1.20.38 pm

Report an Error

Example Question #1 : Analytical Applications Of Derivatives

$\int _0^\pi sin(x) dx$

Possible Answers:

Correct answer:

Explanation:

$\int sin(x) dx = -cos(x) + C$ $\int _0^\pi sin(x) dx = -cos(\pi) - -cos(0)= 1-(-1) = 2$

Report an Error

Example Question #5 : Use Mean Value Theorem, Extreme Value Theorem, And Critical Points

Consider the region bounded by the functions

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

$g(x) = -sin(\frac{\pi x}{2})$

between x = 0 and x = 2 . What is the area of this region?

Possible Answers:

$\frac{4}{3} +\frac{4}{\pi }$

$\frac{4}{3}$

$\frac{3+\pi }{4}$

$\frac{\pi }{4}$

$\frac{3\pi }{4}$

Correct answer:

$\frac{4}{3} +\frac{4}{\pi }$

Explanation:

The area of this region is given by the following integral:

$Area = \int_{0}^{2} f(x) - g(x) = \int_{0}^{2} -x^{2} + 2x - (-sin(\frac{\pi x}{2}))$ or

$Area = \int_{0}^{2} -x^{2} + 2x + sin(\frac{\pi x}{2})$

Taking the antiderivative gives

$-\frac{x^{3}}{3} + x^{2} -\frac{2}{\pi }cos(\frac{\pi x}{2})$ , evaluated from x = 0 to x = 2 .

$F(2) = \frac{-8}{3} + 4 + \frac{2 }{\pi }$ , and

$F(0) = -\frac{2}{\pi }$ .

Thus, the area is given by:

$F(2) - F(0) =- \frac{8}{3} + 4 + \frac{2}{\pi } - (-\frac{2}{\pi }) =\frac{4}{3} +\frac{4}{\pi }$

Report an Error

Example Question #2 : Analytical Applications Of Derivatives

Let $f(x) = x^{4}+ 6x - 17$ .

True or false: As a consequence of Rolle's Theorem, has a zero on the interval (0, 2) .

Possible Answers:

False

True

Correct answer:

False

Explanation:

By Rolle's Theorem, if is continuous on [a,b] and differentiable on (a,b) , and f(a)= f(b) , then there must be $c \in \left ( a,b \right )$ such that f'(c) = 0 . Nothing in the statement of this theorem addresses the location of the zeroes of the function itself. Therefore, the statement is false.

Report an Error

Example Question #7 : Use Mean Value Theorem, Extreme Value Theorem, And Critical Points

$f(x) = x^{3} - 7x + 16$

As a consequence of the Mean Value Theorem, there must be a value $c \in (-2, 2 )$ such that:

Possible Answers:

f'(c) = 3

$f'(c) = \frac{1}{3}$

f'(c) =0

$f'(c) = -\frac{1}{3}$

f'(c) = -3

Correct answer:

f'(c) = -3

Explanation:

By the Mean Value Theorem (MVT), if a function is continuous and differentiable on [a,b] , then there exists at least one value $c \in (a,b)$ such that $f'(c) = \frac{f(b)-f(a)}{b-a}$ . , a polynomial, is continuous and differentiable everywhere; setting a= -2, b= 2 , it follows from the MVT that there is $c \in (-2, 2 )$ such that

$f'(c) = \frac{f(2)-f(-2)}{2-(-2)}$

Evaluating f(-2) and f(2) :

$f(x) = x^{3} - 7x + 16$

$f(2) = 2^{3} - 7 (2) + 16 = 8 -14+16 = 10$

$f(-2) = (-2) ^{3} - 7 (-2) + 16 = -8 +14+16 = 22$

The expression for f'(c) is equal to

$f'(c) = \frac{10-22}{2-(-2)} = \frac{-12}{4} = -3$ ,

the correct choice.

Report an Error

Example Question #8 : Use Mean Value Theorem, Extreme Value Theorem, And Critical Points

is continuous and differentiable on $\mathbb{R}$ .

The values of for five different values of are as follows:

f(0) = 3

f(1) = 4

f(2) = 4

f(3) = 2

f(4) = -1

Which of the following is a consequence of Rolle's Theorem?

Possible Answers:

f(x) must have a zero on the interval (3,4) .

There must be $c \in (1,2)$ such that f'(x) = 0 .

None of the statements in the other choices follows from Rolle's Theorem.

There cannot be $c \in (2,3)$ such that f'(x) = 0 .

f(x) cannot have a zero on the interval (0,3 ) .

Correct answer:

There must be $c \in (1,2)$ such that f'(x) = 0 .

Explanation:

By Rolle's Theorem, if is continuous on [a,b] and differentiable on (a,b) , and f(a)= f(b) , then there must be $c \in \left ( a,b \right )$ such that f'(c) = 0 .

is given to be continuous. Also, if we set a=1, b=2 , we note that f(1) = f(2)=0 . This sets up the conditions for Rolle's Theorem to apply. As a consequence, there must be $c \in (1,2)$ such that f'(c) = 0 .

Incidentally, it does follow from the given information that f(x) must have a zero on the interval (3,4) , but this is due to the Intermediate Value Theorem, not Rolle's Theorem.

Report an Error

Example Question #9 : Use Mean Value Theorem, Extreme Value Theorem, And Critical Points

Find the mean value of the function f(x)=cos(x) over the interval $(0,\frac{\pi}{2})$ .

Possible Answers:

$c=cos\frac{2}{\pi}$

$c=sin^{-1}\frac{2}{\pi}$

$c=sin^{-1}\frac{\pi}{2}$

$c=cos^{-1}\frac{2}{\pi}$

$c=sin\frac{2}{\pi}$

Correct answer:

$c=sin^{-1}\frac{2}{\pi}$

Explanation:

To find the mean value of a function over some interval (a,b) , one mus use the formula: (f(b)-f(a))=(b-a)f'(c) .

Plugging in

$(f(\frac{\pi}{2})-f(0))=(\frac{\pi}{2})(-Sin(c))$

Simplifying

$-1=\frac{\pi}{2}(-Sin(c))$

$\frac{2}{\pi}=Sin(c)$

One must then use the inverse Sine function to find the value c:

$c=sin^{-1}\frac{2}{\pi}$

Report an Error

Example Question #11 : Analytical Applications Of Derivatives

What is a critical value?

Possible Answers:

All of the above

A local min or max

A point of inflection

The point at which the derivative is equal to zero

Correct answer:

All of the above

Explanation:

A point of inflection can be the point at which the derivative is equal to zero, a local minimum or maximum, and a point of inflection. A critical point will always be one that the tangent line to the original function is either completely horizontal or vertical. When the tangent line is completely horizontal, this will be a local maximum or minimum for the function. When the tangent line is completely vertical, then this will be a point of inflection. A point of inflection is a point in which there is a change of curvature in the graph of the function.

Report an Error

Example Question #415 : Calculus Ab

What is the first derivative test in regards to increasing/decreasing intervals?

Possible Answers:

You choose a number less than the critical value. You plug this number into the derivative and if the solution is positive then the function f(x) is increasing, but if the solution is negative then the function f(x) is decreasing.

You choose a number less than, and a number greater than the critical value. You plug these numbers into the derivative and if the solutions are positive, then the function f(x) is increasing but if the solutions are negative then the function f(x) is decreasing.

You choose a number greater than the critical value. You plug this number into the derivative and if the solution is positive then the function f(x) is increasing, but if the solution is negative then the function f(x) is decreasing.

If the critical value is positive, the function f(x) is increasing. If the critical value is negative, the function f(x) is decreasing.

Correct answer:

Explanation:

To find whether a function is decreasing or increasing along an interval, we look at the critical values and use what we call the first derivative test. Take the example f(x) = 2x^2+ 3x . The derivative would be f'(x) = 4x + 3 . To find the critical value we set the derivative equal to zero and solve for .

f'(x) = 4x + 3

0 = 4x + 3

-3 = 4x

$\frac{-3}{4} = x$

Now we have our critical point $\frac{-3}{4}$ . So we choose a number greater than this and plug it back into our derivative function.

f'(0) = 4(0) + 3

f'(0) = 3

The solution is positive. This means that the interval $(\frac{-3}{4}, \infty)$ on the graph of f(x)

f'(-1) = 4(-1) + 3

f'(-1) = -4 + 3 ,

f'(x) = -1

The solution is decreasing. This means that the interval $(-\infty, \frac{-3}{4})$ is decreasing on the graph of f(x) .

Report an Error

← Previous 1 2 3 4 5 6 7 8 Next →

View Tutors

Bianca
Certified Tutor

Central New Mexico Community College, Associate in Science, Diagnostic Medical Sonography.

View Tutors

Morgan
Certified Tutor

York University, Bachelor in Arts, Communication, General. University of Essex, Master of Fine Arts, Directing and Theatrical...

View Tutors

Odysseas
Certified Tutor

University of New Hampshire at Manchester, Master of Science, Civil Engineering. National Technical University of Athens, Doc...

All Calculus AB Resources

45 Practice Tests Question of the Day Flashcards Learn by Concept

Popular Subjects

English Tutors in Dallas Fort Worth, SSAT Tutors in Boston, Statistics Tutors in Miami, SAT Tutors in San Francisco-Bay Area, MCAT Tutors in Los Angeles, ACT Tutors in Philadelphia, Reading Tutors in Miami, Statistics Tutors in Seattle, ISEE Tutors in Houston, SAT Tutors in Washington DC

Popular Courses & Classes

GRE Courses & Classes in Washington DC, GRE Courses & Classes in Seattle, SSAT Courses & Classes in San Diego, GRE Courses & Classes in Atlanta, SAT Courses & Classes in San Diego, LSAT Courses & Classes in Boston, SAT Courses & Classes in Houston, MCAT Courses & Classes in Phoenix, MCAT Courses & Classes in New York City, ISEE Courses & Classes in Boston

Popular Test Prep

SAT Test Prep in San Diego, GMAT Test Prep in Los Angeles, SSAT Test Prep in San Diego, ACT Test Prep in Philadelphia, GMAT Test Prep in Denver, GMAT Test Prep in Washington DC, LSAT Test Prep in Denver, LSAT Test Prep in Miami, ACT Test Prep in Phoenix, GMAT Test Prep in Chicago

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 AB : Analytical Applications of Derivatives

Study concepts, example questions & explanations for Calculus AB

All Calculus AB Resources

Example Questions

Example Question #2 : Use Mean Value Theorem, Extreme Value Theorem, And Critical Points

Example Question #3 : Use Mean Value Theorem, Extreme Value Theorem, And Critical Points

Example Question #1 : Analytical Applications Of Derivatives

Example Question #5 : Use Mean Value Theorem, Extreme Value Theorem, And Critical Points

Example Question #2 : Analytical Applications Of Derivatives

Example Question #7 : Use Mean Value Theorem, Extreme Value Theorem, And Critical Points

Example Question #8 : Use Mean Value Theorem, Extreme Value Theorem, And Critical Points

Example Question #9 : Use Mean Value Theorem, Extreme Value Theorem, And Critical Points

Example Question #11 : Analytical Applications Of Derivatives

Example Question #415 : Calculus Ab

All Calculus AB Resources

Report an issue with this question

DMCA Complaint

Contact Information

Address

Complaint Details

Email address:
Your name:
Feedback: