Mean value theorem: verify g(x)=sqrt(7x-10) satisfies hypothese of MVT

[whitewhimsy]

Verify that g(x)=sqrt(7x-10) satisfies the hypotheses of the Mean Value Theorem on the interval [2,5]. Then, find all numbers c
that satisfy the conclusion of the Mean Value Theorem.

This is what I got so far:

1. g(x) is continuous on [2,5]
2. g(x) is differentiable on [2,5]

g(2)=2
g(5)=5

g'(c)=(g(5)-g(2))/(5-2)
g'(c)=0/0

Isn't this an indeterminate form? I check everything else and they're correct, but I don't understand what to write for my answer and why the Mean Value Theorem doesn't work? Can someone please help me! Any help will be greatly appreciated! Thank you!

 

[ksdhart]

I think you're going to be super embarrassed when you see what you did wrong. According to the Mean Value Theorem, there is at least one c such that:

\(\displaystyle g'\left(c\right)=\frac{g\left(b\right)-g\left(a\right)}{b-a}\)

Plug in the appropriate values a = 2, and b = 5:

\(\displaystyle g'\left(c\right)=\frac{g\left(5\right)-g\left(2\right)}{5-2}=\frac{5-2}{5-2}=\frac{3}{3}=1\)

Then solve for c.

 

[Steven G]

Verify that g(x)=sqrt(7x-10) satisfies the hypotheses of the Mean Value Theorem on the interval [2,5]. Then, find all numbers c
that satisfy the conclusion of the Mean Value Theorem.

This is what I got so far:

1. g(x) is continuous on [2,5]
2. g(x) is differentiable on [2,5]

g(2)=2
g(5)=5

g'(c)=(g(5)-g(2))/(5-2)
g'(c)=0/0

Isn't this an indeterminate form? I check everything else and they're correct, but I don't understand what to write for my answer and why the Mean Value Theorem doesn't work? Can someone please help me! Any help will be greatly appreciated! Thank you!

Talking about hand waving. You need to SHOW in a rigorous manner that g(x) is continuous on [2,5] and differentiable on (2,5).

5-2 is not 0. Be more careful!