Well, first take your derivative. Because derivative gives us the slope at any point. and because maxs and mins have slopes of 0. we would set our derivative to 0.
first: f'(x)=cos(x)+1
so we set that equal to 0
0 = cos(x)+1
and we solve for x.
-1 = cos(x)
arccos(-1) = x
x = pi or 3.14159
so that is our x value at an absolute max or min. you find y values to the left and right of that x value to find out if it is a max or min.
Infinity. that is what I presume
The work you showed is as far as I got. I was under the impression that I should plug the critical points and end points( 0 and pi) back into f(x) to check their values. That would tell me their y value. The end points because it's a closed interval and the critical point because that's where x would be lowest or highest before turning around.
f(0)=0+sin(0)=0
f(pi)=pi+sin(pi)=pi
I was unsure of what to do with those though as it looked weird to me.
For the second one, work would be nice ;o
I played around with some infinity and negative infinity bullstuff. I heard the answer was 1 or something though, not sure yet.
