Derivatives question

**Mr Rayon** · January 15th 2010, 09:50 PM

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.

**Gusbob** · January 15th 2010, 10:03 PM

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.

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