anonymous
  • anonymous
I am so very lost, I believe it dose not apply but I'm unsure. determine whether the mean value theorem to be applied to the function f(x)=2sin(x)+sin2x on the closed interval [7pi,8pi] if the mean value theorem can be applied find all numbers "c" in the open interval (7pi,8pi) such that f'(c)=(f(8pi)-f(7pi))/(8pi-7pi)
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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jamiebookeater
  • jamiebookeater
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zepdrix
  • zepdrix
To be able to apply the Mean Value Theorem, the function must be `continuous` on [7pi,8pi] `differentiable` on (7pi,8pi) So what makes you think it can't be applied? :o The function is certainly continuous, yes? :) It's just the sum of sines.
zepdrix
  • zepdrix
The theorem basically says this: If we have continuity and smoothness of the function, then there is at least one `tangent line` which has the same slope as the `secant line` which connects the end points of our interval
zepdrix
  • zepdrix
Do you understand how to find the slope of that secant line? :)

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anonymous
  • anonymous
Ok
anonymous
  • anonymous
It should be parallel to the tangent line
anonymous
  • anonymous
Draw a line through point A(a(f(a)) and B(b,f(b))
zepdrix
  • zepdrix
|dw:1446923741430:dw|Good yes! :) Same slope = parallel lines. This is an example of what that might look like. In this example we have multiple tangent lines fulfilling this.
anonymous
  • anonymous
Slope is found with f(b)-f(a)/b-a
zepdrix
  • zepdrix
\[\large\rm f(\color{orangered}{x})=2\sin(\color{orangered}{x})+\sin(2\color{orangered}{x}) \]So then,\[\large\rm f(\color{orangered}{8\pi})=2\sin(\color{orangered}{8\pi})+\sin(2\color{orangered}{\cdot8\pi}) \]\[\large\rm f(8\pi)=?\]Can you figure this one?
anonymous
  • anonymous
Hold up I need my calculator
anonymous
  • anonymous
I keep getting 0
zepdrix
  • zepdrix
You shouldn't need a calculator for this :) Gotta get more comfortable with your trig stuff. 8pi and 16pi are multiples of 2pi, which is co-terminal with 0. And sine of 0 is 0. So good, yes. we get 0!\[\large\rm f(8\pi)=0\]\[\large\rm \color{royalblue}{f'(c)=\frac{f(8\pi)-f(7\pi)}{8\pi-7\pi}}\]\[\large\rm \color{royalblue}{f'(c)=\frac{0-f(7\pi)}{8\pi-7\pi}}\]So now how bout \(\large\rm f(7\pi)\) ?
anonymous
  • anonymous
Also 0
anonymous
  • anonymous
|dw:1446924377742:dw|
anonymous
  • anonymous
?
zepdrix
  • zepdrix
k good,\[\large\rm \color{royalblue}{f'(c)=\frac{0-0}{8\pi-7\pi}}\]Simplifying the denominator, \[\large\rm \color{royalblue}{f'(c)=\frac{0}{\pi}}\]And simplifying further,\[\large\rm \color{royalblue}{f'(c)=0}\]
zepdrix
  • zepdrix
It it turns out that this simplifies to an application of Rolle's Theorem! :)
zepdrix
  • zepdrix
We're simply looking for critical points. Those are the `c` values that we want.
zepdrix
  • zepdrix
Take your derivative, what do you get?
anonymous
  • anonymous
The derivative of the original function?
zepdrix
  • zepdrix
Yes.
anonymous
  • anonymous
2cos(x)+sin(2)?
anonymous
  • anonymous
And input 0 for x?
anonymous
  • anonymous
|dw:1446924728113:dw|
zepdrix
  • zepdrix
Hmm that derivative doesn't look right :d
zepdrix
  • zepdrix
\[\large\rm \frac{d}{dx}\sin(2x)\quad=\cos(2x)\frac{d}{dx}(2x)\]You differentiate the out function, sine, giving you cosine. And then chain rule tells you to multiply by the derivative of the inner function.\[\large\rm \frac{d}{dx}\sin(2x)\quad=\cos(2x)\cdot(2)\]\[\large\rm \frac{d}{dx}\sin(2x)\quad=2\cos(2x)\]
zepdrix
  • zepdrix
the outer function* blah typo
anonymous
  • anonymous
|dw:1446924940165:dw|
zepdrix
  • zepdrix
So then,\[\large\rm f(x)=2\sin(x)+\sin(2x)\]\[\large\rm f'(x)=2\cos(x)+2\cos(2x)\]
zepdrix
  • zepdrix
We're interested in f'(x)=0,\[\large\rm 0=2\cos(x)+2\cos(2x)\]
anonymous
  • anonymous
So input 0 as x or just simplify?
zepdrix
  • zepdrix
x is put in for your f'(x). We can't put 0 in for x. We have to solve for x.
anonymous
  • anonymous
|dw:1446925006528:dw|