Take the derivative of F :
F(x) = x³ f(x) + 10 → F'(x) = 3 x² f(x) + x³ f'(x)
The tangent line to F(x) at x = 1 passes through the point (1, F (1)) and has slope equal to F' (1) :
F (1) = 1³ f (1) + 10 = 2
F '(1) = 3 • 1² f (1) + 1³ f' (1) = -12
Use the point-slope formula to find the equation of the line:
y - F (1) = F' (1) (x - 1) → y = -12 x + 14
