Call Now to Set Up Tutoring:

(888) 888-0446

AP Calculus AB : Derivative as a function

Study concepts, example questions & explanations for AP Calculus AB

Create An Account

All AP Calculus AB Resources

3 Diagnostic Tests 164 Practice Tests Question of the Day Flashcards Learn by Concept

Example Questions

Example Question #41 : Relationship Between The Increasing And Decreasing Behavior Of ƒ And The Sign Of ƒ'

Determine the intervals on which the function is increasing:

$f=\frac{x^2}{x-2}, x\neq 2$

Possible Answers:

$(-\infty, -4)$

$(-4, -2)\cup(-2, 0)$

$(0, \infty)$

$(-\infty, -4)\cup(0, \infty)$

Correct answer:

$(-\infty, -4)\cup(0, \infty)$

Explanation:

To determine the intervals on which the function is increasing, we must determine the intervals on which the function's first derivative is positive.

The first derivative of the function is equal to

$f'=\frac{x^2+4x}{(x+2)^2}$

and was found using the following rules:

$\frac{\mathrm{d} }{\mathrm{d} x} \frac{f(x)}{g(x)}=\frac{g(x)f'(x)-f(x)g'(x)}{g(x)^2}$ , $\frac{\mathrm{d} }{\mathrm{d} x} x^n=nx^{n-1}$ , $\frac{\mathrm{d} }{\mathrm{d} x} ax=a$

Next, we must find the critical values, at which the first derivative is equal to zero:

$\frac{x^2+4x}{(x+2)^2}=0$

x(x+4)=0

c=0, -4

Using the critical values, we now create intervals on which to evaluate the sign of the first derivative:

$(-\infty, -4), (-4, -2), (-2, 0), (0, \infty)$

Notice how at the bounds of the intervals, the first derivative is neither positive nor negative, nor is it defined at x=-2 .

Evaluating the sign simply by plugging in any value on the given interval into the first derivative function, we find that on the first interval, the first derivative is positive, on the second and third intervals, the first derivative is negative, and on the fourth interval, the first derivative is positive. Thus, the function is increasing on $(-\infty, -4)\cup(0, \infty)$ .

Report an Error

Example Question #42 : Relationship Between The Increasing And Decreasing Behavior Of ƒ And The Sign Of ƒ'

f'(x)=x^2-3x-36

g'(x)=x^3-7x^2+45

h'(x)=3x^2+4x-130

i'(x)=-x^3+9x^2-3x-24

Which function has a positive slope at x=5?

Possible Answers:

f(x)

i(x)

h(x)

g(x)

None of the other answers.

Correct answer:

i(x)

Explanation:

i(x) is the only function listed with a positive slope at x=5 . The sign of the slope at of a function at a point is determined by the value first derivative of the function at the corresponding point (i.e., f'(x) , g'(x) , h'(x) , and i'(x) ).

i'(5)=-(5)^3+9(5)^2-3(5)-24

=-125+225-15-24=61

61 is positive, and so, i(x) has a positive slope at x=5 . The other derivative functions listed have non-positive outputs at x=5 .

Report an Error

Example Question #371 : Derivatives

Find the area of the region enclosed by the parabola $y=12-x^2$ and $y=-x$ .

Possible Answers:

$\frac{343}{6}$

$\frac{434}{6}$

$\frac{401}{6}$

$\frac{301}{6}$

$\frac{334}{6}$

Correct answer:

$\frac{343}{6}$

Explanation:

The two curves intersect in between x=-3 and x=4 , which can be found by solving the quadratic equation $12-x^2=-x$ .

To solve for the area between curves, y=f(x) and y=g(x) , we use the formula $A=\int^b_a(f(x)-g(x))dx$

For our problem:

$A=\int^4_{-3}(12-x^2-(-x))dx$

$A=12x-\frac{x^3}{3}+\frac{x^2}{2}$ evaluated from to which yields $\frac{343}{6}$ .

Report an Error

Example Question #1 : Mean Value Theorem

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

Possible Answers:

$4$

$2$

$1$

$3$

$5$

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 #1 : Mean Value Theorem

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

Possible Answers:

1.5

3.5

2.5

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 .

Report an Error

Example Question #51 : Derivative As A Function

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

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 #4 : Mean Value Theorem

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:

$\frac{4}{3}$

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

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

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

$\frac{\pi }{4}$

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 #52 : Derivative As A Function

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 #1 : Mean Value Theorem

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

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

Possible Answers:

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

f'(c) = 3

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

f'(c) =0

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 #55 : Derivative As A Function

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

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

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

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.

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

View AP Calculus AB Tutors

Ping
Certified Tutor

The University of Texas at Dallas, Bachelor of Science, Computer Science. University of North Texas, Master of Arts, Education.

View AP Calculus AB Tutors

Alejandro
Certified Tutor

University of Western Ontario, Doctorate (PhD), Mathematics.

View AP Calculus AB Tutors

Jasmine
Certified Tutor

Kansas State University, Bachelor of Science, Computer Science. Ohio State University-Main Campus, Doctor of Science, Mathema...

All AP Calculus AB Resources

3 Diagnostic Tests 164 Practice Tests Question of the Day Flashcards Learn by Concept

Popular Subjects

Chemistry Tutors in New York City, Spanish Tutors in Houston, GRE Tutors in Los Angeles, SSAT Tutors in Miami, English Tutors in Miami, GRE Tutors in Denver, Calculus Tutors in Seattle, English Tutors in Atlanta, SSAT Tutors in San Diego, Math Tutors in Chicago

Popular Courses & Classes

LSAT Courses & Classes in Boston, ISEE Courses & Classes in Seattle, LSAT Courses & Classes in San Francisco-Bay Area, SSAT Courses & Classes in Houston, SAT Courses & Classes in San Diego, LSAT Courses & Classes in Los Angeles, ISEE Courses & Classes in Phoenix, ACT Courses & Classes in New York City, LSAT Courses & Classes in Denver, SAT Courses & Classes in Phoenix

Popular Test Prep

GRE Test Prep in Los Angeles, LSAT Test Prep in Atlanta, GRE Test Prep in Boston, ISEE Test Prep in Miami, SSAT Test Prep in Chicago, ISEE Test Prep in Phoenix, ACT Test Prep in New York City, SAT Test Prep in Washington DC, MCAT Test Prep in Los Angeles, SSAT Test Prep in Washington DC

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

AP Calculus AB : Derivative as a function

Study concepts, example questions & explanations for AP Calculus AB

All AP Calculus AB Resources

Example Questions

Example Question #41 : Relationship Between The Increasing And Decreasing Behavior Of ƒ And The Sign Of ƒ'

Example Question #42 : Relationship Between The Increasing And Decreasing Behavior Of ƒ And The Sign Of ƒ'

Example Question #371 : Derivatives

Example Question #1 : Mean Value Theorem

Example Question #1 : Mean Value Theorem

Example Question #51 : Derivative As A Function

Example Question #4 : Mean Value Theorem

Example Question #52 : Derivative As A Function

Example Question #1 : Mean Value Theorem

Example Question #55 : Derivative As A Function

All AP Calculus AB Resources

Report an issue with this question

DMCA Complaint

Contact Information

Address

Complaint Details

Email address:
Your name:
Feedback: