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

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Answer 1 · 2018-06-12T10:37:51

See process below

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

How do you sketch the graph y=x^4-2x^3+2xy=x4−2x3+2xy=x^4-2x^3+2x using the first and second derivatives?