min and max of f(t) = 2Cost + Sin2t on inverval [0, pi/2]

[member566]

f(t) = 2Cost + Sin2t [0, pi/2]
f is continuous on all real numbers

f'(t) = -2Sint + 2Cos2t =0
2 2

-Sint + Cos2t = 2 f(pi/6) = 2Cospi/6 + Sin2 pi/6
2(2^1/2)/2 + (3^1/2)/2
-Sint + Sin^2t = 0 2^1/2 + 3^1/2
=3
2Sin^2t + sint -1 = 0

(2Sint - 1) (Sint + 1) = 0 f(0) = 2Cos(0) +Sin 2(0)
2(1) + Sin(0) pi
2Sint - 1 = 0 Sint + 1 = 0 2 + 0
=2
2/2Sint = 1/2 Sint = -1

Sint = 1/2 Sint = -1 f(pi/2) = 2Cos(pi/2) + Sin 2 pi/2
2(1/2) + sin(0)
= pi/6 =3pi/2 = 1

Am I missing anything,

Thank you

 

[tkhunny]

How about the Domain?

\(\displaystyle \L\;t\;=\;\frac{\pi}{6}\) is good.

\(\displaystyle \L\;t\;=\;\frac{3}{2}\pi\) is NOT in the Domain

\(\displaystyle \L\;t\;=\;0\) is not considered.

\(\displaystyle \L\;t\;=\;\frac{\pi}{2}\) is not considered.

You have a good start.