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

1 Answer
Nallasivam V
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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