Question

Determine whether the Mean Value Theorem can be applied to f(x) = sqrt (x-7) on the interval [11,23].

If it can be applied, find all the values of c in the interval (11,23) such that f'(c) = f(b) - f(a) / b-a

Answer 1

You have an Order of Operation problem, there. You mean, (f(b)-f(a))/(b-a). This is quite a different animal from what you have written.

What are the qualifications for the application of the MVT?

Answer 2

now this is a lot easier than the last one
i assume this is the last one, but without the typo right?

Answer 3

\[f(b)=f(23)=\sqrt{23-7}=\sqrt{16}=4\]
\[f(a)=f(11)=\sqrt{11-7}=\sqrt{4}=2\]
\[f(b)-f(a)=4-2=2\]
\[\frac{f(b)-f(a)}{b-a}=\frac{4-2}{23-11}=\frac{1}{6}\]

Answer 4

take the derivative, set it equal to \(\frac{1}{6}\) and solve for \(x\)

Answer 5

Mean Value Theorem: Let f be a continuous functions on a compact interval [a,b] that is differential at every point in the interior. Then there exists a point x0 in the interior where f'(x0)=(f(b)-f(a))/(b-a).

\[f(x)=\sqrt{x-7}\]\[\frac{ d }{ dx }f(x)=\frac{ d }{ dx }(\sqrt{x-7})\]\[f'(x)=\frac{ 1 }{ 2 \times \sqrt{x-7} }\]The only place the derivative would not exist is at x=7. However, this is outside the interval, and hence the function is differentiable at every point in the origin.

Also [11,23] is both closed and bounded, so it is compact by definition.

Now\[\frac{ f(b)-f(a) }{ b-a }=\frac{ f(23)-f(11) }{ 23-11 }=\frac{ \sqrt{23-7}-\sqrt{11-7} }{ 12 }=\frac{ \sqrt{16}-\sqrt{4} }{ 12 }=\frac{ 4-2 }{ 12 }=\frac{ 1 }{ 6 }\]

Then\[\frac{ 1 }{ 6 }=\frac{ 1 }{ 2 \times \sqrt{c-7} }\]

Answer 6

Bah, satellite beat me to it. :P