How do you sketch the graph #y=x^3+2x^2+x# using the first and second derivatives?

1 Answer
Jan 26, 2017

Refer Explanation section

Explanation:

Given -

#y=x^3+2x^2+x#

#dy/dx=3x^2+4x+1#

#(d^2y)/(dx^2)=6x+4#

#dx/dy=0=>3x^2+4x+1#

#x=[(-b)+-sqrt(b^2-(4*a*c))]/(2a)#

#x=[(-4)+-sqrt(4^2-(4*3*1))]/(2*3)#

#x=[(-4)+-sqrt(16-(12))]/(6)#

#x=[(-4)+-sqrt(16-12)]/(6)#

#x=[(-4)+-sqrt(16-12)]/(6)#

#x=[(-4)+-sqrt(4)]/(6)#

#x=[(-4)+-2]/(6)#

#x=[-4-2]/(6)=(-6)/6=-1#

#x=[-4+2]/(6)=(-2)/6=-1/3#

At #x=-1#

#(d^2y)/(dx^2)=6(-1)+4=-6+4=-2#

At #x=-1#

#dy/dx=0; (d^2y)/(dx^2)<0#

Hence the function has a maximum.
At #x=-1# the curve is concave downwards.

At #x=-1/3#

#(d^2y)/(dx^2)=6(-1/3)+4=-2+4=2#

At #x=-1#

#dy/dx=0; (d^2y)/(dx^2)>0#

Hence the function has a minimum.
At #x=-1# the curve is concave upwards.

graph{x^3+2x^2+x [-10, 10, -5, 5]}

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