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)=(x1)/(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 don't know how to from here.... can you help...
Q1a) h(x)=√(x+1 ) [3,8]
MVT=[h(b)h(a)]/(ba)=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=[32]/[83] =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) = (32)/(83) = 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) = (x1)/(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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