Find the maxima and minima of function:- #y=2x^3-9x^2-24x+15#?

1 Answer
Evan
May 4, 2018

Minima: #(4,-97)#
Maxima: #(-1,28)#

Explanation:

#y=2x^3-9x^2-24x+15#

#dy/dx=6x^2-18x-24#

Let us determine the coordinates of the maxima and minima,

When #dy/dx=0#,

#6x^2-18x-24=0#

#x^2-3x-4=0#

Factor and solve,

#(x-4)(x+1)=0#

#x=4 or -1#

When #x=4#,

#y=2(4)^3-9(4)^2-24(4)+15#
#color(white)(y)=-97#

#(4,-97)#

When #x=-1#

#y=2(-1)^3-9(-1)^2-24(-1)+15#
#color(white)(y)=28#

#(-1,28)#

Now, to determine the nature of these coordinates,

Find #(d^2y)/dx^2#,

#(d^2y)/dx^2=12x-18#

When #x=4#,

#(d^2y)/dx^2=12(4)-18#
#color(white)((d^2y)/dx^2)=30>0# ( minima )

When #x=-1#,

#(d^2y)/dx^2=12(-1)-18#
#color(white)((d^2y)/dx^2)=-30<0# ( maxima )

Therefore,

#(4,-97)# minima and #(-1,28)# maxima

Check:

graph{2x^3-9x^2-24x+15 [-20, 20, -120,120]}

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