Rolle's Theorem Help.

[revinnova]

Verify that the function satisfies the three hypothesis of Rolle's Theorem on the given interval. Then find all the numbers c that satisfy the conclusion of Rolle's Theorem.

f(x) = sin2pi, [-1,1]

Here's what I've done so far:

-f is continuous on the closed interval [-1,1] because f is a composite function of sine.
-f is differentiable on the interval (-1,1) because 2x is a polynomial.
f(-1) = 0 = f(1).
-f'(c) = cos (2pi c) x 2pi = 2pi cos2pi c = 0
cos2pi c = 0
2pi c = ? stuck here.

 

[galactus]

f(x) = sin2pi, [-1,1]

I am a little confused by this. Do you mean find the point(s) c, in the interval \(\displaystyle (0,2{\pi})\), where sin(x) has derivative 0?. Or the interval [-1,1]?.


f(x)=sin(x) is both continuous and differentiable everywhere. Thus, it is continuous on \(\displaystyle [0,2{\pi}]\) and differentiable on \(\displaystyle (0, 2{\pi})\)

Rolle's theorem says there is at least one point c in \(\displaystyle (0, 2{\pi})\) such that cos(c)=0.

Solving for c yields \(\displaystyle c=\frac{\pi}{2}\;\ and \;\ \frac{3{\pi}}{2}\)

 

[revinnova]

I copied the question directly from the book, and it says the interval [-1,1]. By the way, how do you type the "pi" symbol.

 

[skeeter]

Verify that the function satisfies the three hypothesis of Rolle's Theorem on the given interval. Then find all the numbers c that satisfy the conclusion of Rolle's Theorem.

f(x) = sin2pi, [-1,1]

Here's what I've done so far:

-f is continuous on the closed interval [-1,1] because f is a composite function of sine.
-f is differentiable on the interval (-1,1) because 2x is a polynomial.
f(-1) = 0 = f(1).
-f'(c) = cos (2pi c) x 2pi = 2pi cos2pi c = 0
cos2pi c = 0
2pi c = ? stuck here.

\(\displaystyle \L cos(2\pi c) = 0\)

\(\displaystyle \L 2\pi c = -\frac{3\pi}{2}, -\frac{\pi}{2}, \frac{\pi}{2}, \frac{3\pi}{2}\)

\(\displaystyle \L c = -\frac{3}{4}, -\frac{1}{4}, \frac{1}{4}, \frac{3}{4}\)

edit: fixed my sloppy division

 

[galactus]

Then where does cos c=0 on the interval (-1,1)

\(\displaystyle \frac{\pi}{2}\) is in the interval (-1,1). Where else in that interval does cos(c) = 0?.


EDIT: there ya' go. The skeetmeister took care of you.

 

[revinnova]

Okay, thanks galactus and skeeter. I was thrown off by the variable c, I was use to seeing x. I can't believe I was pondering for 5 mins over that. And yes, it only falls on the points + or - pi/2 and + or - 3pi/2. Once again, thank you for the fast response.