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

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Answer 1 · 2017-05-07T16:54:40

We have a minima at $x=1/2$

We have either a maxima or a minima, when $(dy)/(dx)=0$

As $y=x^2-x$ , $(dy)/(dx)=2x-1=0$ i.e. $x=1/2$

Hence we have a minima or a maxima only at $x=1/2$

Further, if we have $(d^2y)/(dx^2) <0$ , we have a maxima

and if we have $(d^2y)/(dx^2) >0$ , we have a minima

As $(dy)/(dx)=2x-1=0$

$(d^2y)/(dx^2)=2$ and $(d^2y)/(dx^2) >0$

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

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=x^2-x?