Help regarding Rolle's Theorem for f(x)=sin(x/2) [pi/2 ,3pi/2]

[dancerandydance]

First time post here, so here goes.

I'm working on homework, and I have a question that asks that I both verify that the function satisfies the three hypotheses of Rolle's Theorem on the given interval, as well as to find all numbers c that satisfy the conclusion of Rolle's Theorem.

f(x)=sin(x/2) [pi/2 ,3pi/2]

Here's what I've accomplished so far:

f(pi/2) = sin(pi/2/2) = sqrt/2
f(3pi/2) = sin(3pi/2) = sqrt/2
so f(pi/2)=f(3pi/2)=sqrt/2 Which satisfies Rolle's Theorem.

Next,

f'(x)=(cosx/2)/2, which becomes (cosc/2)/2=0

And here is where I fizzle. I'm quite sure I'm just having a major brainfart, but I'm unsure how to go about solving for c from here.

Thank you in advance.

 

[stapel]

I have a question that asks that I both verify that the function satisfies the three hypotheses of Rolle's Theorem on the given interval, as well as to find all numbers c that satisfy the conclusion of Rolle's Theorem.

I'll be working from the statement provided here (a helpful resource, by the way):

Rolle's Theorem:

Suppose f (x) is a function that satisfies all of the following:

. . .1. f (x) is continuous on the closed interval [a, b].

. . .2. f (x) is differentiable on the open interval (a, b).

. . .3. f (a) = f (b).

Then there is a number c such that a < c < b and f '(c) = 0. Or, in other words, f (x) has a critical point in (a, b).

f(x)=sin(x/2) [pi/2 ,3pi/2]

Here's what I've accomplished so far:

f(pi/2) = sin(pi/2/2) = sqrt/2
f(3pi/2) = sin(3pi/2) = sqrt/2
so f(pi/2)=f(3pi/2)=sqrt/2 Which satisfies Rolle's Theorem.

This satisfies the third of the prerequisites. You still need to show (or prove, or state, or something) the continuity and the differentiability of the function. Are you given that sines and cosines are continuous and differentiable, so you can simply state that first two prerequisites are fulfilled? (Probably, yes; but check your notes to be sure.)

Next,

f'(x)=(cosx/2)/2, which becomes (cosc/2)/2=0

How are you getting this? The cosine should not become a cosecant, surely...?

Instead, just work with the trig: You've got y = sin(x/2), so y' = (cos(x/2))*(1/2) = 0.5 cos(x/2). Set this equal to zero. The 0.5 divides off (or the 1/2 multiplies off; either way is the same thing) to give you cos(x/2) = 0. Now apply what you memorized back in trig. (here)

 

[dancerandydance]

Looks like I was a little unclear with my formatting. my cos(x/2)/2 didn't become csc, it became cos(c/2)/2. In the book and other examples, Since I'm finding the c values, they just replace the x with c, so I followed suit. Sorry about the confusion on that part.

 

[stapel]

Okay; well, how far have you gotten with the first two prerequisites of Rolle's Theorem? Where are you stuck in solving the trig equation? Thank you!

 

[dancerandydance]

As far as meeting the three points, it is given that cos is differentiable and continuous. Sorry, left that part out in my original post.

Now with solving the trig function.. I'm drawing a blank. before posting I had made it to cos(x/2)=0. I struggled quite a bit when it came to the trig problems in precalc last semester.
The link you provided is content we covered towards the end of class, and while I did make sense of it, I didn't have a firm enough grasp to memorize it, and I lost my notebook with my notes in a car accident :?. It's just not quite clicking even after reviewing the link.

 

[Deleted member 4993]

As far as meeting the three points, it is given that cos is differentiable and continuous. Sorry, left that part out in my original post.

Now with solving the trig function.. I'm drawing a blank. before posting I had made it to cos(x/2)=0. I struggled quite a bit when it came to the trig problems in precalc last semester.
The link you provided is content we covered towards the end of class, and while I did make sense of it, I didn't have a firm enough grasp to memorize it, and I lost my notebook with my notes in a car accident :?. It's just not quite clicking even after reviewing the link.

cos(x/2)= 0 = cos[(2n+1)*π/2]..... n = 0, ±1, ±2, ±3 .....

x = ??

 

[stapel]

Now with solving the trig function.. I'm drawing a blank. before posting I had made it to cos(x/2)=0. I struggled quite a bit when it came to the trig problems in precalc last semester.

So you followed the link provided in the initial reply, reviewed the topic of graphing sines and cosines (with links from there back to lessons on the trig ratios), and... then what? Where are you still having trouble?

Please be complete. Thank you!

 

[pka]

I'm working on homework, and I have a question that asks that I both verify that the function satisfies the three hypotheses of Rolle's Theorem on the given interval, as well as to find all numbers c that satisfy the conclusion of Rolle's Theorem. f(x)=sin(x/2) [pi/2 ,3pi/2]
f(pi/2) = sin(pi/2/2) = sqrt/2
f(3pi/2) = sin(3pi/2) = sqrt/2
so f(pi/2)=f(3pi/2)=sqrt/2 Which satisfies Rolle's Theorem.
f'(x)=(cosx/2)/2, which becomes (cosc/2)/2=0.

There is but one solution: $\displaystyle c=\pi$

 

[dancerandydance]

Okay, got a chance to come back to this. Did some reading, and this is what I did:

cos(x/2)=0 --> x/2=cos^-1(0), and since I'm taking the inverse of cosine, I use the top half of the unit circle, , which gives me pi/2.
x/2=pi/2, knock out the 2, and I have x=pi, or c=pi. Man I wish I paid a little more attention in precalc.