I know that the max=-b/2a when we have y=ax^2+bx+c but what happens when it's ax^3+bx^2+cx+d?

Question

Please explain in detail
Thank you

Answer 1

I know you mentioned you don't want to use Calculus, but it's really the only way to derive how to get the local min/max.

f(x) = ax^3 + bx^2 + cx + d

f'(x) = 3ax^2 + 2bx + c

0 = 3ax^2 + 2bx + c

x = [-2b +/- sqrt( 4b^2 - 4(3a)(c) )] / (6a)

x = [-2b +/- sqrt (4b^2 - 12ac)] / (6a)

x = [-2b +/- 2sqrt(b^2 - 3ac)] / (6a)

x = [-b +/- sqrt(b^2 - 3ac)] / (3a)

This means we have two values that can possibly be our local min/max. Let's call them x1 and x2.

x1 = [-b + sqrt(b^2 - 3ac)] / (3a) and
x2 = [-b - sqrt(b^2 - 3ac)] / (3a)

All you have to do is calculate these values for x1 and x2, and then plug them into the function f(x), to obtain the potential min or potential max. Note that these will be _local_ min and maxes, since cubics potentially have a range of all real numbers.

Answer 2

You have to use calculus.
Taking the derivative of your function you´ll get the posible points of maxima and minima putting it to zero and studying its sign
f´(x)= 3ax^2 +2bx+c =0 is a second degree equation.
By the way ,if you take the derivative of ax^2 +bx+c which is
2ax+b and put it to zero you´ll get x=-b/2a which is the abscisa of the maximum or minimum.The maximum or minimum is the value of the function at this point