How do you sketch the graph y=x^4-2x^3+2xy=x42x3+2x using the first and second derivatives?

1 Answer
Jun 12, 2018

See process below

Explanation:

1.- Domain is RR
2.- Analyze the roots of equation x^4-2x^3+2x=0 in order to find axis intercepts

Notice that x^4-2x^3+2x=x(x^3-2x^2+2)=0 thus (0,0) is a passing point. There is no more integer roots

3.- First derivative f´(x)=4x^3-6x^2+2=(x-1)^2(4x+2)

Analyze the sign of derivative. (x-1)^2 is allways positive, then the sign of derivative dependes only of sign of 4x+2

4x+2>=0
x>=-1/2

Then function is increasing in (-1/2,oo) and decreasing in (-oo,-1/2) then in x=-1/2 has a minimum

Derivative is zero in x=-1/2 and x=1 but x=1 is not a maximum neither a minimum because

f´´(x)=12x^2-12x=12x(x-1). then f´´ become 0 only in x=0 and x=1. Thus inflection points.

A sketch is presented below
enter image source here