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About Steve Holleran
Expertise
I can help with all math questions from basic math to Calculus. Whether it`s consumer questions, or questions from high school or college students, I have probably dealt with it at some time in my career.

Experience
33 years teaching experience in NJ public schools

Education/Credentials
B.S. Mathematics : Wake Forest University 1972 M.S. Mathematics : Monmouth University 1981

You are here: Experts > Science > Mathematics > Advanced Math > Calculus & Vectors

Topic: Advanced Math

Expert: Steve Holleran
Date: 5/4/2008
Subject: Calculus & Vectors

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

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