Given that the graph f(x) passes through the point (2,3) and that f'(x)=6x^2-14x+3, find f(x).
For this question we already have the derivative of f(x), and so if we integrate it we should get to a general equation with a constant on the end, however we want to get to an exact solution which we should be able to get to by plugging the point into f(x). So it f'(x) = 6x2-14x+3 then f(x) = 6x3/3 - 14x2/2 + 3x/1 + C = 2x3 - 7x2 + 3x + C. So now we have to plug x =2 and f(x) = 3 into this and get 3 = 2 x 8 - 7 x 4 +3 x 2 + C => C = 9. Therefore f(x) = 2x3 - 7x2 + 3x + 9
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