Suppose f(1) = -8 and f'(1) = 12. Let F(x) = x^3f(x)+10. Find an equation of the tangent line to F at x=1. Show all work.

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

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