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

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 Theorem and Rollo’s Theorem.

- Calculus -
**Steve**, Sunday, November 4, 2012 at 12:57am
Rolle'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 1:38am
I 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 1:48am
if 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 1:49am
actually, 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 2:51am
oops. k(0) = -1

adjust the calculation accordingly.

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