# Derivatives question

• Jan 15th 2010, 10:50 PM
Mr Rayon
Derivatives question
Given that $y=\frac{x^3}{3}-x^2+x$, find $\frac{dy}{dx}$. Hence show that $\frac{dy}{dx}\geq 0$ for all x.

Please show complete working out. Any help will be appreciated.
• Jan 15th 2010, 11:03 PM
Gusbob
Quote:

Originally Posted by Mr Rayon
Given that $y=\frac{x^3}{3}-x^2+x$, find $\frac{dy}{dx}$. Hence show that $\frac{dy}{dx}\geq 0$ for all x.

Please show complete working out. Any help will be appreciated.

$y=\frac{x^3}{3}-x^2+x$

$y' = x^2 - 2x + 1$

This can be factorised:
$x^2 - 2x + 1 = (x-1)^2$

Since y' is a square number, it has to be zero or positive for all x.