Wednesday

September 17, 2014

September 17, 2014

Posted by **Daniella** on Saturday, November 3, 2012 at 11:57pm.

Q1a) h(x)=√(x+1 ) [3,8]

Q1b) K(x)=(x-1)/(x=1) [0,4]

Q1c) Explain the difference between the Mean Value Theorem and Rollo’s Theorem.

- Calculus -
**Steve**, Sunday, November 4, 2012 at 12:57amRolle's theorem is just the mean value theorem, where f(x) = 0 at both endpoints.

What did you get to parts a and b?

- Calculus -
**Daniella**, Sunday, November 4, 2012 at 1:38amI get stuck at this and dont know how to from here.... can you help...

Q1a) h(x)=√(x+1 ) [3,8]

MVT=[h(b)-h(a)]/(b-a)=h'(c)

To find h(b) and h(a), we just plug endpoints into original function

h(b)=h(8)=√(x+1 )

h(b)=h(8)=√(8+1 ) = 3

h(a)=h(3)= √(3+1 ) = 2

MVT=[3-2]/[8-3] =f^' (c)

MVT=1/2=f^' (c)

Next, we find the derivative for h(x)

h'(x)=√(x+1 )

h'(x)=(d/dx(x+1))/(2√(x+1))

h'(x)=(1+0)/(2√(x+1))

h'(x)=1/(2√(x+1))

h'(c)=h'(x)

1/2=(1+0)/(2√(x+1))

- Calculus -
**Steve**, Sunday, November 4, 2012 at 1:48amif h(x) = √(x+1)

h'(x) = 1/(2√(x+1))

so, we want c where

h'(c) = (3-2)/(8-3) = 1/5

1/2√(x+1) = 1/5

5 = 2√(x+1)

√(x+1) = 5/2

x+1 = 25/4

x = 21/4

and 3 < 21/4 < 8

-----------------------------

if k(x) = (x-1)/(x+1)

k'(x) = 2/(x+1)^2

k(0) = 0

k(4) = 3/5

so, we want c where k'(c) = (3/5)/4 = 3/20

3/20 = 2/(x+1)^2

3(x+1)^2 = 40

(x+1)^2 = 40/3

x = -1 + 2√(10/3) = 2.65

0 < 2.65 < 4, so we're ok.

- Calculus -
**Steve**, Sunday, November 4, 2012 at 1:49amactually, I think Rolle's Theorem is the MVT where f(a) = f(b), so that f'(c) = 0.

- Calculus -
**Steve**, Sunday, November 4, 2012 at 2:51amoops. k(0) = -1

adjust the calculation accordingly.

**Answer this Question**

**Related Questions**

calculus - verify that the function satisfies the hypothesis of the mean value ...

math - verify that the function satisfies the hypothesis of the mean value ...

calculus - Verify that the Intermediate Value theorem applies to the indicated ...

Calculus - Verify that the hypotheses of the Mean-Value Theorem are satisfied ...

Caluclus - [Mean Value Theorem] f(x)=-3x^3 - 4x^2 - 2x -3 on the closed interval...

math - Verify that f(x) = x^3 − 2x + 6 satisfies the hypothesis of the ...

math - very urgent ! - Verify that f(x) = x^3 − 2x + 6 satisfies the ...

calculus - Verify that the hypotheses of the Mean-Value Theorem are satisfied on...

math - verify that the function satisfies the hypotheses of the mean values ...

calculus - Let f(x) = (x+1)/(x-1). Show that there are no vlue of c such that f(...