# How do you find all local maximum and minimum points using the second derivative test given y=x^2-x?

May 7, 2017

We have a minima at $x = \frac{1}{2}$

#### Explanation:

We have either a maxima or a minima, when $\frac{\mathrm{dy}}{\mathrm{dx}} = 0$

As $y = {x}^{2} - x$, $\frac{\mathrm{dy}}{\mathrm{dx}} = 2 x - 1 = 0$ i.e. $x = \frac{1}{2}$

Hence we have a minima or a maxima only at $x = \frac{1}{2}$

Further, if we have $\frac{{d}^{2} y}{{\mathrm{dx}}^{2}} < 0$, we have a maxima

and if we have $\frac{{d}^{2} y}{{\mathrm{dx}}^{2}} > 0$, we have a minima

As $\frac{\mathrm{dy}}{\mathrm{dx}} = 2 x - 1 = 0$

$\frac{{d}^{2} y}{{\mathrm{dx}}^{2}} = 2$ and $\frac{{d}^{2} y}{{\mathrm{dx}}^{2}} > 0$

hence we have a minima at $x = \frac{1}{2}$ at which $y = \frac{1}{4} - \frac{1}{2} = - \frac{1}{4}$

graph{x^2-x [-2.126, 2.874, -1.09, 1.41]}