Question

find the x-coordinate of all points on the curve y=12Xcos(5X)-(30*sqrt3*X^2) + 16, pi/5<X<2pi/5 where the tangent line passes through the point (0,16), (not on the curve)

I have absolutely no idea how to solve this one

Answer 1

If I read your formula correctly,

f(x) = 12x cos(5x)-(30√3 x^2) + 16

The slope at any point on the curve is

f'(x) = 12(cos5x - 5x(sin5x + √3))
So, the equation of the tangent line at x=h is

y-f(h) = f'(h)(x-h)
y-(12h cos(5h)-(30√3 h^2) + 16) = (12(cos5h - 5h(sin5h + √3)))(x-h)

Since this line passes through (0,12), we have

12-(12h cos(5h)-(30√3 h^2) + 16) = (12(cos5h - 5h(sin5h + √3)))(-h)

Hmmm. wolframalpha gets no real solutions for that. Looking at the graph, it does appear that the tangent might pass through (0,12) at about x=0.8

The tangent lines near there are
at x=0.8, y = 20.19-54.66x
at x=0.9, y = 10.58-42.27x

So there should be a solution at about 0.88 or so. Hmmm.

Maybe I made a typo in here somewhere.