Find absolute extrema on the closed interval...?

Question

Find absolute extrema of the function g(x) = 2x + 5cosx on the interval [0,2pi]

I was trying to find the critical numbers, and I ended up with sinx = 2/5 and I'm not quite sure what (rational) value of x satisfies that.

I'm trying to find the critical numbers, so I could find the y value at them for g, and compare them to the y values at the endpoints, to find the absolute extrema.

Anyone could tell me what I'm doing wrong? Or what I should be doing next?

(Note: just an answer to the question will be useless to me, I need to understand how to do the question, otherwise I have the answer in the back of my textbook).

Thanks a lot.

Answer 1

f'(x)=2-5sinx
setting this to zero
sinx=2/5
absolute extreme=2(2/5)+cosarcsin(2/5)

Answer 2

Setting f'(x)=0 gets you relative extrema, but you need to check the endpoints of the interval, too.

So d(2x + 5cos x)/dx = 2 - 5 sin x = 0 when sin x = 2/5, x = arcsin 2/5, x in [0,2π] = .4115 AND x = π - .4115 = 2.7301.

So you consider those 2 points AND the endpoints of the interval.
f(0) = 5
f(.4115) = 5.406
f(2.7301) = .8776
f(2π) = 17.5664

and from that the absolute extrema are obvious.

By the way, there's NO reason to expect your x's to be rational. At best, your first answerer could have worked out cos(arcsin 2/5) = (√21)/5, but he didn't mention that over the interval [0, 2π] arcsin x = 2/5 has TWO solutions, something that should be obvious from the graph of sin x. The definition of inverse sine is restricted to [-π/2, π/2] so that it will be a function.

Answer 3

Take the first derivative and find where it's equal to zero and when it's undefined. These are your critical points. Check to the left and right to see if the function is increasing or decreasing, or just take the second derivative and decide if it's concave up or down. This will give you your extrema.