What are the extrema of #y=2x^3 - 5x^2 - 4x + 7#?

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Answer 1 · 2018-04-13T18:08:43

#x=-1/3# local maximum, #x=5/6# is turning point and #x=2# local minimum.

After taking derivative of #y# and equating to #0#,

#6x^2-10x-4=0#

#(3x+1)*(2x-4)=0#, so #x1=-1/3# and #x_2=2#

For #x=-1/3#, #y=140/27#. So #x=-1/3# local maximum.

For #x=2#, #y=-7#. So #x=2# local minimum.

After taking second derivative of #y# and equating to #0#,

#12x-10=0#, so #x=5/6# is turning point.