Asked by chrissy

find the absolute maximum and minimum for f(x) = x+ sin(x) on the closed interval [0, 2pi]

Answers

Answered by Steve
f' = 1 + cos(x)
f'=0 when cos(x) = -1, or x=pi

however,
f''(x) = -sin(x) which is also 0 at x=pi.

So, (pi,pi) is an inflection point, not a max or min.

f(2pi) = 2pi

so that is the absolute max on [0,2pi]

Note that f'(x) >= 0 for all x, so there is no max or min, which require changing direction.
There are no AI answers yet. The ability to request AI answers is coming soon!

Related Questions