Question

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

Answer 1

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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