Please help me with this question

Question

Consider the function g(x) = (f (− cos(x)))^3 .

The function f(x) is an unknown, but differentiable function. We know the following values for f(x) and f '(x).

x	f(x)	f'(x)
-√2/2	0	2
-1/2	-7	-9
-√3/2	2	-route(5)

Find the equation of the tangent line to g(x) at x = π/6

Answer 1

The slope-intercept formula is y = mx + b, where m is the slope and b is the y-intercept. For the tangent line at π/6, the formula can be rewritten as y = g'(π/6)x + b

g(x) = (f(-cos(x)))^3

g'(x) = 3(f(-cos(x)))^2 * f'(-cos(x)) * sin(x)

g'(π/6) = 3(f(-cos(π/6)))^2 * f'(-cos(π/6)) * sin(π/6) = 3(f(-√3/2))^2 * f'(-√3/2) * 1/2 = 3(2)^2 * -route(5) * 1/2 = -6route(5)

I'm going to assume route(5) means 5, so g'(π/6) = -30

Rewriting the formula from step 1, y = g'(π/6)x + b, we get y = -30x + b

To find the second unknown, b, we need to get the point (π/6, g(π/6)).

g(π/6) = (f(-cos(π/6)))^3 = (f(-√3/2))^3 = 2^3 = 8.

We can now solve for b using the point (π/6, 8)

8 = (-30 * (π/6)) + b

8 = -5π + b

8 + 5π = b

The tangent line to g(x) at x = π/6 is y = -30x + 8 + 5π.