1. ## extremum

Find all extrema in the interval $[0,2{\pi}]$if y=x + sin x.

$f'(x)=1+cos(x)$

$1+cos(x)=0$

What solutions to this fall in your interval?.

Here's a graph. They always help.

3. is it

a) $(-1,-1+ \frac {3\pi} {2}), (0,0)$
b) $(2 {\pi}, 2{\pi})$ (0,0)
c) $(2{\pi}, 2{\pi}), ({\pi},{\pi})$
d) $({\pi},{\pi}),$(0,0)
e) None of these

4. Originally Posted by Samantha
is it

a) $(-1,-1+ \frac {3\pi} {2}), (0,0)$
b) $(2 {\pi}, 2{\pi})$ (0,0)
c) $(2{\pi}, 2{\pi}), ({\pi},{\pi})$
d) $({\pi},{\pi}),$(0,0)
e) None of these
Have you looked at Galactus's plot? Are there any local maxima or minima?
Where are the global maximum and minimum in $[0,2 \pi ]$?

RonL

5. Originally Posted by CaptainBlack
Have you looked at Galactus's plot? Are there any local maxima or minima?
Where are the global maximum and minimum in $[0,2 \pi ]$?

RonL
I'mnot sure =(

6. I'm sorry to say, you should see your professor if you're that lost.

7. That's how I did it and understand it. Plz help

y=x+sinx
y(prime)=1+cosx
cosx=-1
x= $\pi$

so the answer it $(\pi, \pi)$

8. Originally Posted by Samantha
That's how I did it and understand it. Plz help

y=x+sinx
y(prime)=1+cosx
cosx=-1
x= $\pi$

so the answer it $(\pi, \pi)$
Evidently from the graph the function only has extrema at the ends of the interval since it is monotonically increasing. So the extrema are at x = 0 and x = 2*pi.

-Dan

9. Originally Posted by topsquark
Evidently from the graph the function only has extrema at the ends of the interval since it is monotonically increasing. So the extrema are at x = 0 and x = 2*pi.

-Dan
Then the answer is $(2\pi, 2\pi),$(0,0)?

10. Originally Posted by Samantha
Then the answer is $(2\pi, 2\pi),$(0,0)?
Looks like it. The critical points within the interval are obviously not relative max or min points (which you can tell by graphing or the second derivative test, whichever you prefer.)

-Dan