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

Question

1401 views around the world

You can reuse this answer
Creative Commons License

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]}