How to Determine the Coefficients of a Cubic Function with Given Conditions?

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To determine the coefficients of the cubic function f(x) = ax^3 + bx^2 + cx + d given the conditions of relative maximum at (3,3), relative minimum at (5,1), and an inflection point at (4,2), the first derivative must equal zero at x=3 and x=5. This implies that f'(x) can be expressed as a quadratic function of the form a(x-3)(x-5). The second derivative must equal zero at the inflection point x=4, providing another equation. By substituting these points into the first and second derivatives, three equations can be established to solve for the coefficients a, b, c, and d. This methodical approach will yield the explicit form of the cubic function.
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Homework Statement


Find a,b,c, and d such that the cubic f(x)=ax^{3}+bx^{2}+cx+dsatisfies the indicated conditions.

Relative maximum (3,3)
Relative minimum (5,1)
Inflection point (4,2)

Homework Equations





The Attempt at a Solution




I am so lost as to how to do this :/.

Its a polynomial so f ' (x) must = 0 at x=3 and x=5 (can't not exist), and I also know that the derivative of f(x) will be a function of degree 2, which can have at most two roots. Thus the function must be of the form a(x-3)(x-5)=f ' (x), right?

I know that the second derivative is defined for all x (can't have negative exponents, they would become constants before that point). And that f '' (x)=0 at x=4...

I just can't see how to piece it all together. Can someone help me out?
 
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Work out the explicit form of the first and second derivatives of f. Then plug in x=3 and x=5 into the first derivative, which you said must be zero. Plug in x=4 into the second derivative, which again, you said must be zero. This will give you 3 equations in 3 unknowns, which you can solve.
 
f'(x)=3a(x-3)(x-5)\equiv 3ax^2 +2bx +c

is a place to start.
 
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