Consider the following function for t ≥ 0, y=exp(-3*t)*cos(kx-9*t+0.8)

(a) The derivative dy/dt (at fixed k and x) =

(b) At t=0, the maximum value of y =

(c) At t=0, the minimum value of y =

(e) For x=0, at what t closest to zero is dy/dt=0? t =

(6,1)(5,4)slope = (4 - 1) / (5 - 6) = 3/-1 = -3y = mx + bslope(m) = -3use either of ur , = 6 and y = 1now sub and find b, the y int1 = -3(6) + b1 = -18 + b1 + 18 = b19 = bso ur equation is : y = -3x + we need it in standard formy = -3x + 3x to both sides3x + y = 19 < == standard form