How Do I solve this

Sin2πx [-1,1]

Yes continuous

Yes differentiable

f(-1) = f(1)

So now derive

2cos2πx = 0

How do I solve this? The answer in my book says +/- 1/4, +/- 3/4

How

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- Oct 15th 2012, 05:49 PMskinsdomination09How Do I solve this (using rolle's)?!?! (sin2pi*x) on -1,1 interval(TESTPREP PLZHELP)
How Do I solve this

Sin2πx [-1,1]

Yes continuous

Yes differentiable

f(-1) = f(1)

So now derive

2cos2πx = 0

How do I solve this? The answer in my book says +/- 1/4, +/- 3/4

How - Oct 15th 2012, 05:59 PMTheEmptySetRe: How Do I solve this (using rolle's)?!?! (sin2pi*x) on -1,1 interval(TESTPREP PLZH
I am going to have to GUESS what the question is....

I am assuming that you are trying to find the points in $\displaystyle c \in [-1,1] $ where $\displaystyle f'(c)=0$

If so you need to solve the equation

$\displaystyle f'(x)=2\pi \cos(2\pi x)=0 \iff \cos(2\pi x) =0$

Taking the arccosine of both sides gives

$\displaystyle 2\pi x =\cos^{-1}(0) \iff 2 \pi x =\frac{\pi}{2} +\pi n, \quad n \in \mathbb{Z}$

$\displaystyle x=\frac{1}{4}+\frac{n}{2}, \quad n \in \mathbb{Z}$

This will generate all of the solutions to the problem.