Expert: Steve Holleran - 5/4/2008

Question
Find all maximum and minimum points for y=2sinx+sin^2x

Answer
Hi Nazia,

Okay, so we have y = 2 sin x + sin^2 x

So the derivative is  y' = 2 cos x + 2 (sin x )(cos x) =0

                       = 2 cos x (1 + sin x) = 0

                           cos x = 0 or   1 + sin x = 0

 Now, cos x = 0 at x = pi/2, 3pi/2,... = pi/2 +/- k*pi
 (all pi multiples of pi/2).

1 + sin x = 0 when sin x = -1, and this is x = 3*pi/2, but we've already accounted for that in the solution to cos x = 0.

A sign chart on y' would be :


               -             +           -            +
   y'  --------------|-------------|----------|-------
                   -pi/2         pi/2        3pi/2

so it looks like you get maximums at pi/2 +/- 2k*pi and minimums at
3pi/2 +/- 2k*pi.

The max would be y = 2 sin(pi/2) + sin^2(pi/2) = 2 + 1 = 3

The min would be y = 2 sin(3pi/2) + sin^2(3pi/2) = -2 + 1 = -1


Hope this is what you needed.

Steve