two hard problems?

Question

Find the absolute maximum value f on the given interval.
f(t) = 2cos(t) + sin(2t)
[0, pi/2]

Find the absolute maximum value of f on the given interval.
f(x) = e^(-x) - e^(-8x)
[0, 1]

Answer 1

f´(t) --2sin t+2 cos2t=0
-2sin t +2(1-2 sin ^2 t) = 0
sin t=z
4z^2+2z -2=0 // 2z^2+z-1=0
z=((-1+-3))/4 so z= -1 and z= 1/2
sin t = -1 t = 3pi/2 outside the interval
sin t= 1/2 t= pi/6
f(0) = 2, f(pi/6) = sqrt(3) +sqrt(3)/2 ,f(pi/2) =0
The absolute maximum is sqrt(3)(3/2)
2)
f´(x) = -e^-x+8e^-8x = e^-x( -1+e^-7x)
e^-7x=1 so x= 0
f(0) =0,f(1) = 1/e-1/e ^8 absolute maximum