# extremum

• Jul 30th 2007, 04:00 AM
Samantha
extremum
Find all extrema in the interval$\displaystyle [0,2{\pi}]$if y=x + sin x.
• Jul 30th 2007, 04:15 AM
galactus

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

$\displaystyle 1+cos(x)=0$

What solutions to this fall in your interval?.

Here's a graph. They always help.
• Jul 30th 2007, 04:46 AM
Samantha
is it

a)$\displaystyle (-1,-1+ \frac {3\pi} {2}), (0,0)$
b) $\displaystyle (2 {\pi}, 2{\pi})$ (0,0)
c) $\displaystyle (2{\pi}, 2{\pi}), ({\pi},{\pi})$
d) $\displaystyle ({\pi},{\pi}),$(0,0)
e) None of these
• Jul 30th 2007, 04:54 AM
CaptainBlack
Quote:

Originally Posted by Samantha
is it

a)$\displaystyle (-1,-1+ \frac {3\pi} {2}), (0,0)$
b) $\displaystyle (2 {\pi}, 2{\pi})$ (0,0)
c) $\displaystyle (2{\pi}, 2{\pi}), ({\pi},{\pi})$
d) $\displaystyle ({\pi},{\pi}),$(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 $\displaystyle [0,2 \pi ]$?

RonL
• Jul 30th 2007, 05:04 AM
Samantha
Quote:

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 $\displaystyle [0,2 \pi ]$?

RonL

I'mnot sure =(
• Jul 30th 2007, 05:07 AM
galactus
I'm sorry to say, you should see your professor if you're that lost.
• Jul 31st 2007, 11:40 AM
Samantha
That's how I did it and understand it. Plz help

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

so the answer it $\displaystyle (\pi, \pi)$
• Jul 31st 2007, 12:41 PM
topsquark
Quote:

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=$\displaystyle \pi$

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

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
• Aug 1st 2007, 04:18 AM
Samantha
Quote:

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 $\displaystyle (2\pi, 2\pi),$(0,0)?
• Aug 1st 2007, 04:31 AM
topsquark
Quote:

Originally Posted by Samantha
Then the answer is $\displaystyle (2\pi, 2\pi),$(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