Suppose that f has a continuous second derivative for all x, and that f(0)=1, f'(0)=2, and f''(0)=0.

Question

A. Does f have an inflection point at x=0?
B. Let g'(x) = (3x^2 + 2)f(x) + (x^3 + 2x + 5)f'(x). The point (0,5) is on the graph of g. Write the equation of the tangent line to g at this point.
C. Use your tangent line to approximate g(0.3).

Answer 1

A. since f'≠0 and f"=0, that is an inflection point
B. by the chain rule, g = (x^3+2x+5)*f(x)
g'(0) = 2f(0) + 5f'(0) = 2+10 = 12
so now you have a point and a slope, so the tangent line is
g-5 = 12(x-0)
C. ∆g ≈ g' * ∆x
so add that to g(0), with ∆x = 0.3

Answer 2

Thank you!
How do I find g''(0)?