Thread: Show the cubic function has extreme value

1. Show the cubic function has extreme value

Show that the cubic $\displaystyle p(x) = x^3 + ax^2 + bx + c$ has extreme values iff $\displaystyle a^2 > 3b.$

2. cubics

Any cubic goes from -infinityto +infinity. The extreme values occur when there is a bump, i.e., the derivative is zero. When you take the derivative, you'll get a quadratic. If it has real roots, you've found where the bumps are. The quadratic formula should help you.

3. Originally Posted by qmech
Any cubic goes from -infinityto +infinity. The extreme values occur when there is a bump, i.e., the derivative is zero. When you take the derivative, you'll get a quadratic. If it has real roots, you've found where the bumps are. The quadratic formula should help you.
ok, so now I have $\displaystyle p'(x) = 3x^2 + 2ax +b$

so do I set p'(x) = 0 and find the root? But how do I solve the equation?

4. Use the quadratic formula. You know, that stuff with $\displaystyle x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$

5. Originally Posted by 450081592
ok, so now I have $\displaystyle p'(x) = 3x^2 + 2ax +b$

so do I set p'(x) = 0 and find the root? But how do I solve the equation?
$\displaystyle p'(x)$ needs to have two real roots ... the discriminant, $\displaystyle b^2-4ac > 0$

$\displaystyle (2a)^2 - 4(3)(b) > 0$

$\displaystyle 4a^2 - 12b > 0$

$\displaystyle a^2 - 3b > 0$

$\displaystyle a^2 > 3b$