Given the function f(x) = x^3 - 5x^2 + 3x + a, find the conditions at which f(x) > 0 is always true for any positive value of x

Answer 1

A=9.

differentiating x^3 - 5x^2 + 3x

3x^2 - 10x + 3

the min occurs at ( 3x - 1)(x - 3 ) = 0

x=1/3 and x=3

(1/3)^3 - 5(1/3)^2 + 3(1/3) = 1/27 - 5/9 + 1 = (1-15 + 27)/27 = 13/27

3^3 - 5*3^2 + 3*3 = 27 - 45 + 9 = -18 + 9 = -9

So the graph needs shifted UP 9 points.

A=9