Question

Determine whether the Mean Value Theorem, Applies to
the following functions on the given interval [a,b].
(b) If so, find or approximate the point(s) that are
guaranteed to exist by the MEan Value Theorem.
f(x) = 3sin(2x); [0,pi/4]

Answer 1

Would I not be able to use the MVT because the domain
on the closed interval [0,pi/4] is every number except
at 0?

Answer 2

Making the function not qualify for being continuous according to the rules of continuity?

Answer 3

3 sin(2x) is continous

Answer 4

Oh wait, I forgot that the domain is all values that x can be without making the function undefined.
So now that I have that I can use the MVT to get:
\[\frac{3\sin(\pi/2)-3\sin(0)}{\pi/4-0}=\frac{3}{\pi/4}=12/\pi \]
So I need to proove that f prime at c is 12/pi.

Answer 5

If the function is continuous over the closed interval and differentiable over the open interval, then the MVT says there *is* a value of \(c\) between \(0\) and \(\dfrac{\pi}{4}\) such that
\[f'(c)=\frac{f\left(\dfrac{\pi}{4}\right)-f(0)}{\dfrac{\pi}{4}-0}\]
You have to find this \(c\).

Answer 6

\[f(x)=3\sin2x~~\Rightarrow~~f'(x)=6\cos2x~~\Rightarrow~~f'(c)=6\cos2c\]
So you solve for \(c\):
\[6\cos2c=\frac{f\left(\dfrac{\pi}{4}\right)-f(0)}{\dfrac{\pi}{4}-0}\]

Answer 7

I know you could get rid of the 6 to get:
\[=>\cos(2c)=2/\pi\]
From there I'm not sure. Could I get rid of the 2 to get:
\[=>\cos(c)=1/\pi\]

Answer 8

\[\begin{align*}6\cos2c&=\frac{f\left(\dfrac{\pi}{4}\right)-f(0)}{\dfrac{\pi}{4}-0}\\\\
&=\frac{3\sin\dfrac{2\pi}{4}-3\sin0}{\dfrac{\pi}{4}}\\\\
&=\frac{3-0}{\dfrac{\pi}{4}}\\\\
6\cos2c&=\frac{12}{\pi}\\\\
\cos2c&=\frac{2}{\pi}\\\\
2c&=\cos^{-1}\left(\frac{2}{\pi}\right)\\
c&=\frac{1}{2}\cos^{-1}\left(\frac{2}{\pi}\right)
\end{align*}\]
You'll need a calculator for this one

Answer 9

How do you go from:
\[\cos2c=2/\pi \]
to
\[=>2c=\arccos(2/\pi)\]

Answer 10

the inverse cos "undoes" the cos

if you have
cos(x) = a
and take the inverse cosine of both sides:
\[ \cos^{-1}(\cos(x)) = \cos^{-1}(a) \\ x= \cos^{-1}(a) \]

Answer 11

oh I see, Thanks.