Minima Maxima: y=ax³+bx²+cx+d

What is the condition that the cubic y=ax³+bx²+cx+d shall have two Extremes? Answer nya po is b²-3ac>0 Salamat po.

Differentiate the given cubic equation and equate to zero. Solve the roots of the resulting quadratic equation. To have two extremes, the roots of the quadratic equation must be real. What makes the roots real is when the discriminant of the quadratic equation greater than zero.

If you have further question, don't hesitate to ask.

$y = ax^3 + bx^2 + cx + d$

$y' = 3ax^2 + 2bx + c = 0$
 

The roots of the quadratic equation are
$x = \dfrac{-B \pm \sqrt{B^2 - 4AC}}{2A}$

$x = \dfrac{-2b \pm \sqrt{4b^2 - 12ac}}{6a}$
 

To have two extremes
$4b^2 - 12ac \gt 0$

When i substitute x to y", answer nya po is 2√b²-3ac and -2√b²-3ac?

Pero yung answer daw po is only b²-3ac.
How po?

In reply to by Francis June E…

The problem ask a condition for the cubic to have two extremes. You give "4b2 - 12ac > 0" as your condition to satisfy the problem.
 

From $x = \dfrac{-2b \pm \sqrt{4b^2 - 12ac}}{6a}$

  • If $4b^2 - 12ac \lt 0$, x is imaginary. (no extreme)
  • If $4b^2 - 12ac = 0$, you only have one value of x. (one extreme)
  • If $4b^2 - 12ac \gt 0$, there are two values of x. (two extremes)

 

That is why you give $4b^2 - 12ac \gt 0$ as you answer to satisfy the condition of the problem.