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
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
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.