what is a cubic function with a local maximum value of 60 at 1 and a local minimum value of -4 at 3

Question

i need help finding a cubic function with a local max of 60 at 1 and a local min of -4 at 3

Answer 1

Let the function be f(x)=ax^3+bx^2+cx+d, then f'(x)=3ax^2+2bx+c=0 at maximum or minimum.

But we know the solutions are x=1 and 3: p(x-1)(x-3)=0=px^2-4px+3p; thus c=3p; a=p/3; b=-2p.

f(x)=px^3/3-2px^2+3px+d, where p and d have yet to be found.

(1,60) and (3,-4) must lie on the curve, so f(1)=60 and f(3)=-4:

60=p/3-2p+3p+d=4p/3+d; -4=9p-18p+9p+d, d=-4 and 4p/3-4=60; 4p/3=64; p=48, a=16, b=-96, c=144.

f(x)=16x^3-96x^2+144x-4=4(4x^3-24x^2+36x-1).