JAMB - Mathematics (2006 - No. 11)
Find the value of x for which the function 3x\(^3\) - 9x\(^2\) is minimum
zero
2
3
5
Giải thích
y = 3x\(^3\) - 9x\(^2\)
\(\frac{dy}{dx}\) = 9x\(^2\) - 18x
As \(\frac{dy}{dx}\) = 0
9x\(^2\) - 18x = 0
9x(x-2) = 0
9x = 0 which implies x = 0
x - 2 = 0 implies x = 2
\(\frac{d^2y}{dx^2}\) = 18x - 18
when x = 0
\(\frac{d^2y}{dx^2}\) > 0 ∴ x is minimum
when x = 2, \(\frac{d^2y}{dx^2}\) = 18
∴ the value > 0 x is minimum
\(\frac{dy}{dx}\) = 9x\(^2\) - 18x
As \(\frac{dy}{dx}\) = 0
9x\(^2\) - 18x = 0
9x(x-2) = 0
9x = 0 which implies x = 0
x - 2 = 0 implies x = 2
\(\frac{d^2y}{dx^2}\) = 18x - 18
when x = 0
\(\frac{d^2y}{dx^2}\) > 0 ∴ x is minimum
when x = 2, \(\frac{d^2y}{dx^2}\) = 18
∴ the value > 0 x is minimum