Question

Please help?
consider the function on the interval (0,2pi) For each function, (a) find the open interval(s) on which the function is increasing or decreasing, (b) apply the First Derivative Test to identify all relative extrema

f(x)=x+2 sin x

Answer 1

HOw do I do this with the interval?

Answer 2

Start normally, but only look at 0 to 2pi.

Answer 3

I know the derivative is f'(x)=2 cos (x) + 1

Answer 4

what do I do from there?

Answer 5

Hi, Vibes,

Start by finding the first derivative of f(x) = x + 2sin x. You've just done this correctly.

Next step is to set this derivative = to 0 and to solve for x; any solutions you find on the interval (0, 2pi) are "critical values."


Graph those critical values on a number line which starts at 0 and ends at 2pi.

Create subintervals on the number line based upon these critical values.

Choose a test number from each subinterval and determine whether the derivative f'(x) is + or - on that interval.

Can you take this problem from there?

Answer 6

so set 2 cos (x) + 1 =0

Answer 7

how do I cancel out the cos?

Answer 8

2cos x +1 =0 --> 2cos x = -1--> cos x = -1/2 so far so good?

Answer 9

oh ok I see

Answer 10

is that for the (o, 2 pi)

Answer 11

There are several directions in which you could go from here. You could draw a diagram showing the two triangles within a circle of radius 2 for which the cosine of the central angle is -1/2. Or you could use the inverse cosine function, arccos x, applying it to -1/2: arccos (-1/2). Or you could graph cos x on (0, 2pi) and determine the angles at which this function is = to -1/2.

Answer 12

I don't know how to do any of that. :( ...