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

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Answer 1 · 2017-05-07T16:54:40

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

We have either a maxima or a minima, when $\frac{d y}{d x} = 0$ $(dy)/(dx)=0$

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

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

Further, if we have $\frac{d^{2} y}{{d x}^{2}} < 0$ $(d^2y)/(dx^2) <0$ , we have a maxima

and if we have $\frac{d^{2} y}{{d x}^{2}} > 0$ $(d^2y)/(dx^2) >0$ , we have a minima

As $\frac{d y}{d x} = 2 x - 1 = 0$ $(dy)/(dx)=2x-1=0$

$\frac{d^{2} y}{{d x}^{2}} = 2$ $(d^2y)/(dx^2)=2$ and $\frac{d^{2} y}{{d x}^{2}} > 0$ $(d^2y)/(dx^2) >0$

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

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

How do you find all local maximum and minimum points using the second derivative test given y=x^2-xy=x2−xy=x^2-x?