Given that $\displaystyle y=\frac{x^3}{3}-x^2+x $, find $\displaystyle \frac{dy}{dx}$. Hence show that $\displaystyle \frac{dy}{dx}\geq 0$ for all x. Please show complete working out. Any help will be appreciated.
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Originally Posted by Mr Rayon Given that $\displaystyle y=\frac{x^3}{3}-x^2+x $, find $\displaystyle \frac{dy}{dx}$. Hence show that $\displaystyle \frac{dy}{dx}\geq 0$ for all x. Please show complete working out. Any help will be appreciated. $\displaystyle y=\frac{x^3}{3}-x^2+x $ $\displaystyle y' = x^2 - 2x + 1 $ This can be factorised: $\displaystyle 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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