# Calculus (Intermediate Value Theorem)

39,001 results

**calculus**

Verify that the Intermediate Value theorem applies to the indicated interval and find the value of c guaranteed by the theorem. f(x) = x^2 - 6x + 8, [0,3], f(c) = 0 I have no idea how to use the theorem :(

**Math Calculus**

The Image Theorem: The image theorem, a corollary of the intermediate value theorem, expresses the property that if f is continuous on the interval [a, b], then the image (the set of y-values) of f on [a,b] is all real numbers between the minimum of f(x) on [a,b], inclusive. ...

**Math - Calculus**

Show that the equation x^3-15x+c=0 has at most one root in the interval [-2,2]. Perhaps Rolle's Theorem, Mean Value Theorem, or Intermediate Value Theorem hold clues? ...Other than simply using my TI-84, I have no idea how to accomplish this.

**Math - Calculus**

Show that the equation x^3-15x+c=0 has at most one root in the interval [-2,2]. Perhaps Rolle's Theorem, Mean Value Theorem, or Intermediate Value Theorem hold clues? ...Other than simply using my TI-84, I have no idea how to accomplish this.

**Calculus**

Use the intermediate value theorem to find the value of c such that f(c) = M. f(x) = x^2 - x + 1 text( on ) [-1,12]; M = 21

**intermediate value thorem**

Use intermediate value theorem to show that the polynomial function has a zero in the given interval. f(-3)= value of 0=

**calculus**

verify the Intermediate Value Theorem if F(x)=squre root of x+1 and the interval is [3,24].

**calculus**

Use the Intermediate Value Theorem to show that there is a root in the equation x^(1/3)=1-x in the interval (0,1).

**calculus**

"use the intermediate value theorem to prove that the curves y=x^2 and y=cosx intersect"

**Calculus**

use the intermediate value theorem to determine whether there is a zero f(x) = -3^3 - 6x^2 + 10x + 9 ; [-1,0]

**Calculus**

Verify the hypothesis of the mean value theorem for each function below defined on the indicated interval. Then find the value “C” referred to by the theorem. Q1a) h(x)=√(x+1 ) [3,8] Q1b) K(x)=(x-1)/(x=1) [0,4] Q1c) Explain the difference between the Mean Value ...

**Calculus (Intermediate Value Theorem)**

If f(x)= x^3-x+3 and if c is the only real number such that f(c)=0, then c is between ______?

**Calculus**

Determine whether the hypotheses of the Intermediate-Value Theorem are satisfied. f(x)=x^1/3 , [a,b]=[-1,1] Please, explain. Thank you.

**Calculus**

Suppose f(x) = x ^ 4 – 4x ^ 2 + 6, and g(x) = 3x ^ 3 – 8x. Prove, via the Intermediate Value Theorem, that the functions intersect at least twice between x = –2 and x = 4.

**calculus**

Use the intermediate value theorem to determine whether or not f(x)=x^2+7x-7 and g(x)=4x+21 intersects on [-4,-1]. If applicable, find the point of intersection on the interval.

**Use the intermediate value**

Use the intermediate value theorem to show that the polynomial function has a zero in the given interval. f(x)=9x^4-3x^2+5x-1;[0,1]

**Use the intermediate value**

Use the intermediate value theorem to show that the polynomial function has a zero in the given interval. f(x)=4x^3+6x^2-7x+1; [-4,-2] f(-4)=

**College Algebra**

1. Use the Intermediate Value Theorem to show that the polynomial function has a zero in the given interval. f(x) = 13x^4 - 5x^2 +7x -1; [3,0] Enter the value of (-3). 2. Use the Intermediate Value Theorem to show that the polynomial function has a zero in the given interval. ...

**Calculus**

Let f be a twice-differentiable function such that f(2)=5 and f(5)=2. Let g be the function given by g(x)= f(f(x)). (a) Explain why there must be a value c for 2 < c < 5 such that f'(c) = -1. (b) Show that g' (2) = g' (5). Use this result to explain why there must be a ...

**Math**

Use the Intermediate Value Theorem to show that there is a root of the given equation in the specified interval. cos x = x. How do I begin this problem? According to the theorem, a=0, b=1 and N=x?

**Calculus**

Consider the function f(x)=8.5x−cos(x)+2 on the interval 0¡Üx¡Ü1 . The Intermediate Value Theorem guarantees that there is a value c such that f(c)=k for which values of c and k? Fill in the following mathematical statements, giving an interval with non-zero length ...

**calculus**

Q: Suppose that for all xE(0,5), f(x) is between 1+x and 3+sin((pi)(x)). Find lim x->2 f(x). Is this question related to the intermediate value theorem? It is confusing me, can anyone help out? I am not certain of what xE(0,5) is defining, is that a set of x values or am I ...

**Calculus (Please Check)**

Show that the equation x^5+x+1 = 0 has exactly one real root. Name the theorems you use to prove it. I.V.T. *f(x) is continuous *Lim x-> inf x^5+x+1 = inf >0 *Lim x-> -inf x^5+x+1 = -inf <0 Rolles *f(c)=f(d)=0 *f(x) is coninuous *f(x) is differentiable f'(x) = 5x^4...

**Calculus**

Sorry... Consider the function f(x) = 8.5 x − cos(x) + 2 on the interval 0 ¡Ü x ¡Ü 1. The Intermediate Value Theorem guarantees that there is a value c such that f(c) = k for which values of c and k? Fill in the following mathematical statements, giving an interval ...

**Math**

Use the Intermediate Value Theorem to show that there is a root of the given equation in the specified interval. x^4+x-3=0, interval (1,2). According the to theorem, I found that a is 1, b is 2 and N is 0. f(1)= 2 and f(2) = 17. Is the root (1.16,0)?

**Calculus**

Given f(x) = -1/x, find all c in the interval [-3, -½] that satisfies the Mean Value Theorem. A. c= -sqrt(3/2) B. c= +or- sqrt(3/2) C. The Mean Value Theorem doesn’t apply because f is not continuous at x=0 D. The Mean Value Theorem doesn’t apply because f(-½) does not ...

**calculus**

Use the Intermediate Value Theorem to prove that the equation has a solution. Then use a graphing calculator or computer grapher to solve the equation. 2x^3-2x^2-2x+1=0 i am completely lost & have no idea where to start.

**Algebra**

Use the intermediate theorem to show that the polynomial function value has a zero in the given interval f(x)=x^5-x^4+8x^3-7x^2-17x+7; [1.6,1.8] Find the value of f(1.6) Find the value of f(1.8)

**Calculus**

Use the Fundamental Theorem of Calculus to find the area of the region bounded by the x-axis and the graph of y = 4 x3 − 4 x. Answer: (1) Use the Fundamental Theorem of Calculus to find the average value of f(x) = e0.9 x between x = 0 and x = 2. Answer: (2) Draw the ...

**math**

use intermediate value theorem to show f(x) has a zero f(x)= x^5 - 4x^4- 7x^2 - 6x; [-0.7, -0.6]

**math**

use intermediate value theorem to show f(x) has a zero f(x)= x^5 - 4x^4- 7x^2 - 6x; [-0.7, -0.6]

**Math**

Use intermediate value theorem to show there is a root to 2x^3 + x^2 - 2 = 0 on [0,1]

**calculus**

Let f(x) = (x+1)/(x-1). Show that there are no vlue of c such that f(2)-f(0) =f'(c)(2-0). Why does this not contradict the Mean Value Theorem? I plugged 2 and 0 into the original problem and got 3 and -1 . Then I found the derivative to be ((x-1)-(x+1))/(x-1)^2. Whould would I...

**Calculus**

Let f(x)=αx^2+βx+γ be a quadratic function, so α≠0, and let I=[a,b]. a) Check f satisfies the hypothesis of the Mean Value Theorem. b)Show that the number c ∈ (a,b) in the Mean Value Theorem is the midpoint of the interval I.

**calculus**

verify that the function satisfies the hypothesis of the mean value theorem on the given interval. then find all numbers c that satisfy the conclusion of the mean value theorem. f(x) = x/(x+2) , [1,4]

**Calculus**

show that ((x − 1)/x) <( ln x) < (x − 1) for all x>1 Hint: try to apply the Mean Value Theorem to the functions f(x) = lnx and g(x) = xlnx. I'm having trouble applying the mean value theorem

**math**

Use the intermediate value theorem to verity that x^4+X-3=0 has a solution in the interval(1,2)

**calculus**

Referring to the Mean Value Theorem and Rolle's Theorem, how can I tell if f is continuous on the interval [a,b] and differentiable on (a,b).

**Economic**

1) GDP does not include intermediate goods because a. that would understate the true size of GDP. b. intermediate goods are not useful to consumers. c. that would count the value of intermediate goods twice. d. intermediate goods are not valuable. 2) The dollar value of an ...

**Calculus I**

Suppose that f and g are two functions both continuous on the interval [a, b], and such that f(a) = g(b) = p and f(b) = g(a) = q where p does not equal to q. Sketch typical graphs of two such functions . Then apply the intermediate value theorem to the function h(x) = f(x) - g...

**Algebra**

For f(x) = x^3 – 4x – 7, use the Intermediate Value Theorem to determine which interval must contain a zero of f.

**precalulus**

use the intermediate value theorem to show that f(x) has a zero in the given interval. f(x) = -x^5 -2x^4 + 5x^3 + 4; [-0.9, -0.8] Stuck!

**Calculus**

use the intermediate value theorem to prove that every real number has a cubic root. That is, prove that for any real number a there exists a number c such that c^3=a

**AP Calculus**

Show that the equation x^3 - 15x + c = o has exactly one real root. All I know is that it has something to do with the Mean Value Theorem/Rolle's Theorem.

**Calculus**

Determine if the Mean Value Theorem for Integrals applies to the function f(x) = 5 - x^2 on the interval 0 to sqrt 5 . If so, find the x-coordinates of the point(s) guaranteed by the theorem.

**Calculus**

Does F(x)=9-x^2 satisfy the hypothesis of the mean value theorem on the interval [0,3]? if it does, then find the exact values of all Ce(0,3) that satisfy the conclusion of the Mean Value Theorem.

**calculus**

usethe intermediate theorem to show that the polynomial function has a zero in the given interval f(x)=18x^4-8x^2+9x-1;[0,3) can you please me how you got the answer

**college algebra**

Use the intermediate value theorem to determine whether the polynomial function has a zero in the given interval. f(x)=8x^5-4x^3-9x^2-9;[1,2]

**calculus**

Consider the function f(x)=65x−cos(x)+2 on the interval 0 less than or equal to x less than or equal to 1. The Intermediate Value Theorem guarantees that there is a value c such that f(c)=k for which values of c and k? Fill in the following mathematical statements, ...

**calculus**

Verify that the hypotheses of the Mean-Value Theorem are satisfied on the given interval, and find all values of c in that interval that satisfy the conclusion of the theorem. f(x)=x^2-3x; [-2,6]

**Calculus**

Verify that the hypotheses of the Mean-Value Theorem are satisfied for f(x) = √(16-x^2 ) on the interval [-4,1] and find all values of C in this interval that satisfy the conclusion of the theorem.

**calculus**

determine whether the mean value theorem can be applied to f on the closed interval [a,b]. If the Mean Value Theorem can be applied, find all values of c in the open interval (a,b) such that f(c) =f(b) - f(a) / b - a

**calculus**

Determine if the Mean Value Theorem for Integrals applies to the function f(x) = x^3 − 9x on the interval [−1, 1]. If so, find the x-coordinates of the point(s) guaranteed to exist by the theorem

**algebra**

use the intermediate value theorem to show the polynominal function has a zero in the given interval f(x)=x^5-x^4+3x^3-2x^2-11x+6; [1.5,1.9] x= -2.33 y=10.19 after i plugged in the 1.5 and 1.9 i just want to know if my x and y are correct

**Math**

On which interval does the Intermediate Value Theorem guarantee that the polynomial x^4 + 7x^2 − 9x − 1 has a root? A. (-1/2,0) B. (1/2,1) C. (0,1/2) D. (-1,-1/2)

**Math**

On which interval does the Intermediate Value Theorem guarantee that the polynomial x^4 + 7x^2 − 9x − 1 has a root? A. (-1/2,0) B. (1/2,1) C. (0,1/2) D. (-1,-1/2)

**Math**

On which interval does the Intermediate Value Theorem guarantee that the polynomial x^4 + 7x^2 − 9x − 1 has a root? A. (-1/2,0) B. (1/2,1) C. (0,1/2) D. (-1,-1/2)

**Calculus**

Determine if the Mean Value Theorem for Integrals applies to the function f(x) = x^3 - 9x on the interval [-1, 1]. If so, find the x-coordinates of the point(s) guaranteed to exist by the theorem. I think one point is zero.

**Math**

Let f(x) = 2x + 1 − sin(x), how many roots does f(x) have in the interval [−π, π]? Use the next steps to prove that it has only one root. a) Use the Intermediate Value Theorem to show that f(x) has at least one root. (b) Explain why f(x) is increasing on ...

**Calculus**

For f(x)=x^2/3(x^2-4) on [-2,2] the "c" value that satisfies the Rolle's Theorem is A. 0 B. 2 C. +or-2 D. There is no value for c because f(0) does not exist E. There is no value for c because f(x) is not differentiable on (-2,2)

**college algebra--need help please!!**

use the intermediate value theorem to show that the polynomial function has a zero in the given interval. f(x)=x^5-x^4+9x^3-5x^2-16x+5;[1.3,1.6] f(x)1.3= ? simplify answer f(x)1.6= ? "

**CALCULUS!**

suppose that 3 <_ f prime of x <_ 5, for all values x. show that 18<_ f(8)-f(2) <_ 30 <_ signs mean less or equal to... im supposed to apply mean value theorem or rolle's theorem... i don't understand neither so i cant do the question! please help!

**algebra**

Show Work Use the intermediate value theorem to show that the polynomial function has a zero in the given interval. f(x)=x^5-x^4+4x^3-4x^2-20x+18 [1.5,1.8]

**math**

Q: Suppose that for all xE(0,5), f(x) is between 1+x and 3+sin((pi)(x)). Find lim x->2 f(x). Is this question related to the intermediate value theorem? It is confusing me, can anyone help out? I am not certain of what xE(0,5) is defining, is that a set of x values or am I ...

**Calculus**

Consider the function f(x)=6x-cos(x)+5 on the interval 0 is less than or equal to x, and x is less than or equal to 1. The Intermediate Value Theorem guarantees that there is a value c such that f(c)=k for which values of c and k? Fill in the following mathematical statements...

**calculus**

Determine if the Mean Value Theorem for Integrals applies to the function f(x) = x^3 − 9x on the interval [−1, 1]. If so, find the x-coordinates of the point(s) guaranteed to exist by the theorem. Not quite sure how to do this one at all to be completely honest. ...

**calculus**

Use the intermediate value theorm to show that the polynomial function has a zero in the given interval f(x)=x^5-x^4+8x^3-5x^2-14x5;[1.4;1.5] find the value of f(1.4) f(1.4)= find the value of f(1.5) f(1.5)=

**calculus**

let f(x)= (x-3)^-2 Show that there is no value of c in (1,4) such that f(4)-f(1)= (f prime of c)(4-1). Why doesn't this contradict the mean value theorem.

**Calculus I Theorem**

I factored and simplified dy/dx of 192x^5 + 96x^3 + 12x all the way down to x^2 = u = (-1/2) and (-1/6). How does the result illustrates part 1 of the Calculus Fundamental Theorem?

**calculus**

let f(x)= 2 - |2x-1|. Show that there is no value of c such that f(3)- f(0) = f'(c)(3-0). Why does this not contradict the mean value theorem.

**Calculus**

Find the biggest value of c that satisfy the Mean Value Theorem for integrals for f(x)= 1/(x+1)^6 on the interval [0,7]

**Calc**

Use the Intermediate Value Theorem to check whether the equation x^3–3x+2.1=0 has a root in the interval (0,1) answer: yes or no ? i have no idea how to answer to go about solving this question, thanks for the help!

**Calculus**

Find the value(s) of c guaranteed by the Mean Value Theorem for Integrals for the function over the given interval. f(x) = 4√x [4, 9]

**calculus help**

Does the function satisfy the hypotheses of the Mean Value Theorem on the given interval? f(x)= ln(x) , [1,6] If it satisfies the hypotheses, find all numbers c that satisfy the conclusion of the Mean Value Theorem. (Enter your answers as a comma-separated list. If it does not...

**Calculus Help Please!!!**

Does the function satisfy the hypotheses of the Mean Value Theorem on the given interval? f(x) = 2x^2 − 5x + 1, [0, 2] If it satisfies the hypotheses, find all numbers c that satisfy the conclusion of the Mean Value Theorem. (Enter your answers as a comma-separated list...

**Calculus Help Please!!!**

does the function satisfy the hypotheses of the Mean Value Theorem on the given interval? f(x) = 2x^2 − 5x + 1, [0, 2] If it satisfies the hypotheses, find all numbers c that satisfy the conclusion of the Mean Value Theorem. (Enter your answers as a comma-separated list...

**calculus**

Find the value or values of c that satisfy the equation f(b)-f(a)/b-a = f'(c) in the conclusion of the mean value theorem for the given function and interval f(x)= x^(2/3) , [0,1]

**calculus**

Verify the means value theorem holds on the interval shown. Then, find the value c such that f'(c)=(f(b)-f(a))/(b-a) a. f(x)= x-1/x on [1,3] b.f(x)=x^3=x-4 on [-2,3] c. f(x)= x^3 on [-1,2] d. f(x)= Sqr. root of x on [0,4]

**Algebra**

For f(x) = x^3 – 4x – 7, use the Intermediate Value Theorem to determine which interval must contain a zero of f. A. Between 0 and 1 B. Between 1 and 2 C. Between 2 and 3 D. Between 3 and 4

**Algebra**

For f (x) = x4 – 2x2 – 7, use the Intermediate Value Theorem to determine which interval Must contain a zero of f. A. Between 0 and 1 B. Between 1 and 2 C. Between 2 and 3 D. Between 3 and 4

**Math**

Use the intermediate value theorem to show that f(x) has a zero in the given interval. Please show all of your work. f(x) = 3x^3 + 8x^2 - 5x - 11; [-2.8, -2.7] I'm not understanding all this questions...

**Calculus--Pythagorean Theorem**

Use the Pythaogorean Theorem to determine the exact length of AB. Express the answer as A) an exact value in simplest mixed radical form B) A decimal to the nearest hundredth The picture is right here, I uploaded it of the diagram. h t t p : //imageshack . us/photo/my-images/...

**college algebra, Please help!!**

use the intermediate value theorem to show that the polynomial function has a zero in the given interval. f(x)=4x^3+3x^2-8x+7;[-5,-2] please show all work

**calculus**

use the intermediate value theoremto show that the polynomial function has a zero in the given interval f(x)=x^5-x^4+8x^3-5x^2-14x5;[1.4;1.5] find the value of f (1.4)= find the value of f (1.5)= show work, thanks

**Calculus**

Apply mean value theorem f(x)=7-(6/x) on [1,6]

**Calculus**

use mean value theorem: f(x)= 7- 2^x, [0,4], c=?

**Calculus**

Determine if the Mean Value Theorem for Integrals applies to the function f(x) = x^3 − 16x on the interval [−1, 1]. If so, find the x-coordinates of the point(s) guaranteed to exist by the theorem. f(a)=15 f(b)=-15 f'(x)=3x^2-16 f'(c)=-15 I am now lost from here, I do not ...

**Calculus**

I finished the first question with no problem. It was something like: Find the points at which f(x) [insert three equations for left right and in between here] is discontinuous. At each of these points, is f cont. from the right or the left? As I said, I solved that one. I am ...

**algebra**

Want to make sure I am doing this correctly. Use the Intermediate Value Theorem for polynomials to show that the polynomial function f(x)has zero between the numbers given. f(x)=3x^2-x-4 ; 1 and 2 3(1)^2-1-4= - 2 below zero 3(2)^2-1-4= 6 above zero Am I doing this problem correct

**Caluclus**

[Mean Value Theorem] f(x)=-3x^3 - 4x^2 - 2x -3 on the closed interval [0,8]. Find the smallest value of c that satisfies the conclusion of the Mean Value Theorem for this function defined on the given interval. I got 8 - sqrt(5696) / -18 = 3.748436059 but it's not right.

**AP CALCULUS**

Using the mean value theorem; F'(x) = f(b)-f(a) / b-a f(x)=x^2-8x+3; interval [-1,6]

**Calculus I**

Section The fundamental Theorem of Calculus: Use Part I of the fundamental Theorem to compute each integral exactly. 4 | 4 / 1 + x^2 dx 0

**Algebra**

For f(x) = x^3 – 4x – 7, use the Intermediate Value Theorem to determine which interval must contain a zero of f. A. Between 0 and 1 B. Between 1 and 2 C. Between 2 and 3 D. Between 3 and 4 I am leaning towards Choice A. What does everyone think? I would appreciate some ...

**math**

verify that the function satisfies the hypothesis of the mean value theorem on the given interval. then find all numbers c that satisfy the conclusion of the mean value theorem. f(x) = x/(x+2) , [1,4]

**Calculus**

Use Mean Value Theorem and find all numbers c in (a,b) 1.x+(4/x) [1,4] Help me!

**Calculus**

Find the number c that satisfies the conclusion of the Mean Value Theorem. f(x) = x/(x + 4) [1, 8]

**Calculus**

Use the Evaluation Theorem to find the exact value of the integral ç 7 1 1/5x(dx)

**Calculus**

Find the values of c guaranteed by the mean value theorem for integrals. f(x)= x^3 [0,3]

**Calculus**

Find all values of c that satisfy the Mean Value Theorem for f(x) = x^3 + 1 on [2, 4].

**Calculus**

The function defined below satisfies the Mean Value Theorem on the given interval. Find the value of c in the interval (1, 2) where f'(c)=(f(b) - f(a))/(b - a). f(x) = 1.5x-1 + 1.1 , [1, 2] Round your answer to two decimal places.