Minima Maxima: y=ax³+bx²+cx+d

14 posts / 0 new
Last post
Francis June E....
Francis June E. Seraspe's picture
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.

Jhun Vert
Jhun Vert's picture

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.

Francis June E....
Francis June E. Seraspe's picture

D ko po makuha yung roots sir

Jhun Vert
Jhun Vert's picture

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

$y' = 3ax^2 + 2bx + c = 0$
 

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
$4b^2 - 12ac \gt 0$

Francis June E....
Francis June E. Seraspe's picture

Yan na po value ni x?

Francis June E....
Francis June E. Seraspe's picture

Paano po makuha c 4b²-12ac?

Jhun Vert
Jhun Vert's picture

Substitute:
A = 3a
B = 2b
C = c

Francis June E....
Francis June E. Seraspe's picture

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?

Jhun Vert
Jhun Vert's picture

You don't need to do that. Divide 4b2 - 12ac > 0 both sides by 4 and you will get the answer.

Francis June E....
Francis June E. Seraspe's picture

Pero paano po nakuha yung 4b²-12ac?

Francis June E....
Francis June E. Seraspe's picture

When i substitute A,B and C

The answer is -2b+-√4b²-12ac/6a

Paano po naging 4b²-12ac nalang po?

Francis June E....
Francis June E. Seraspe's picture

D ko parin po makuha kung paano naging b²-3ac nalang

Jhun Vert
Jhun Vert's picture

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.
 

Francis June E....
Francis June E. Seraspe's picture

Okay na po. Salamat po .

 
 
Subscribe to MATHalino on