1. ## Rolles Theorem

I am having trouble with this problem

Use a graphing utility to graph the function on the closed interval [a, b].
f (x) =(x/2)-sin(pi*x/6) [−1, 0]

Determine whether Rolle's Theorem can be applied to f on the interval.
If Rolle's Theorem can be applied, find all values of c in the open interval (a, b) such that f '(c) = 0. (Enter your answers as a comma-separated list. Round your answers to four decimal places. If not possible, enter IMPOSSIBLE.)

2. Right skewed arc. Just graph it with a calculator if you wan to see it.

3. Originally Posted by sydewayzlocc
can you explain why it doesnt work
$\frac{dy}{dx}\bigg[\frac{x}{2}-sin\bigg(\frac{\pi x}{6}\bigg)\bigg]=\frac{1}{2}-\frac{\pi cos\big(\frac{\pi x}{6}\big)}{6}$

Set equal to 0

$\frac{1}{2}-\frac{\pi cos\big(\frac{\pi x}{6}\big)}{6}=0\rightarrow x=\frac{6arccos\big(\frac{3}{\pi}\big)}{\pi}$

4. Originally Posted by dwsmith
Still doesn't work.
this still doesnt make sense, i tried saying that it was impossible and it said wrong answer, I need more help with this problem

5. $f'(c)=0$

$\frac{1}{2}-\frac{\pi cos\big(\frac{\pi x}{6}\big)}{6}=0\rightarrow x=\frac{6arccos\big(\frac{3}{\pi}\big)}{\pi}$

6. Originally Posted by sydewayzlocc
I am having trouble with this problem

Use a graphing utility to graph the function on the closed interval [a, b].
f (x) =(x/2)-sin(pi*x/6) [−1, 0]

Determine whether Rolle's Theorem can be applied to f on the interval.
If Rolle's Theorem can be applied, find all values of c in the open interval (a, b) such that f '(c) = 0. (Enter your answers as a comma-separated list. Round your answers to four decimal places. If not possible, enter IMPOSSIBLE.)
What trouble are you having? Drawing the graph with a graphing utility? Do you know what Rolles Theorem says? Are you having trouble:

Determining if Rolles Theorem can be applied? (Can you evaluate f(-1)? Can you evaluate f(0)? Can you see whether or not f is continuous over the given interval?) Can you differentiate f? Can you solve the resulting equation f'(x) = 0?

It is no good saying "I'm having trouble". You need to show what you've been able to do and then say where you're having trouble.