Minima Maxima: y=ax³+bx²+cx+d

What is the condition that the cubic y=ax³+bx²+cx+d shall have two Extremes?

Salamat po.

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Differentiate the given cubic

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.

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D ko po makuha yung roots sir

D ko po makuha yung roots sir

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

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

$y' = 3ax^2 + 2bx + c = 0$

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

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Yan na po value ni x?

Yan na po value ni x?

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Paano po makuha c 4b²-12ac?

Paano po makuha c 4b²-12ac?

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

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When i substitute x to y",

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?

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You don't need to do that.

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

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Pero paano po nakuha yung

Pero paano po nakuha yung 4b²-12ac?

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When i substitute A,B and C

When i substitute A,B and C

Paano po naging 4b²-12ac nalang po?

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D ko parin pa makuha kung

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

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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.

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Paano ko po malalaman na 4b²

Okay na po. Salamat po .

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