You are not "dumping" homework on us, are you?
I will do the first one, you do the others and let me know what you got
a)f(x) = -cos 3x - 4sin 3x
f'(x) = 3sin(3x) - 12cos(3x)
the slope of the tangent when x=π/6
= 3sin(π/2) - 12cos(π/2)
= 3 - 0 = 3
so the slope of the normal is -1/3
when x=π/6
f(π/6) = -cos π/2 - 4sin π/6 = 0 - 4 = -4
We need the equation of a line with slope -1/3 and a point (π/6, -4) on it
using y = mx + b
-4 = (-1/3)π/6 + b
b = π/18 - 4 or (π - 72)/18
y = (-1/3)x + π/18 - 4 or y = (-1/3)x + (π-72)/18
weloo.
Consider the function f(x)= -cos3x -4sin3x.
(a)Find the equation of the line normal to the graph of f(x) when x= pie/6 .
(b)Find the x coordinates of the points on the graph of f(x) where the tangent to the graph is horizontal.
(c)Find the absolute extrema of the function f(x)=5+6x^3-3x^4 on the interval [-2,2] .
1 answer