Question

A function of the form y = csin(bt2) whose first critical point for positive t occurs at t = 1 and whose derivative is 7 when t = 2.

Answer 1

7 = 2bc(2)cos(4b)
7 = 4bccos(4b)
i get to this point then im lost

Answer 2

now i have \[\frac{ 7 }{ 4\pi }= \cos(4\pi/c)\]