I've been stuck on this particular question for two days now:
f(x)=e^2x sin2x 0 ≤ x ≤ pi
a) Use calculus to find the coordinates of the turning points on the graph of y=f(x).
b) Show that f''(x)=8e^2x cos2x
c) Hence, or otherwise, determine which turning point is a maximum and which is a minimum.
I've only managed part of a )and only a little of b) Here's my working so far:
a) f'(x)= e^2x times 2cos2x + 2e^2x times sin2x
f'(x)= 2e^2x(cos2x) + 2e^2x(sin2x)
2e^2x( sin2x + cos2x )=0
2e^2x=0 or sin2x + cos2x=0
sin2x = -cos2x So divide both sides by cos2x
So tan2x = -1. That's all I've been able to do so far for a), I have no idea what to do next as the particular book I'm woking on only explains the derivates and intergration of sin and cos but not tan. I've understood sin/cos = tan because of a previous question also resulting in tan and connecting the dots. But I have absolutely no idea what to do from there, I just know sin/cos=tan.
In regards to b): f'(x)= 2e^2x( sin2x + cos2x )
So f''(x)= 2e^2x times ( 2cos2x -2sin2x ) + 4e^2x times ( sin2x + cos2x ). I have tried to simplify from there but it's no use since I have no idea how I'm supposed to prove the answer is 8e^2x cos2x.
I haven't even attempted to start c) yet.
Thank you to anyone who took the time to read this and can enlighten me. Calculus can sometimes be very confusing/irritating!