How do you graph the inequality $y>=x^3-6x^2+12x-8$ ?

Question

1771 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2017-09-10T06:43:56

Please see below.

Note that considering the equality $x=2$ , $f(x)=x^3-6x^2+12x-8=0$ , hence $(x-2)$ is a factor of $x^3-6x^2+12x-8$

$x^3-6x^2+12x-8$

= $(x-2)(x^2-4x+4)$

= $(x-2)^3$

Observe that for $x=2$ , we have $y=0$ and hence the curve $y=(x-2)^3$ just moves the function $y=x^3$ , $2$ points to the right.

The graph of $(x-2)^3=0$ appears as follows:

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

This divides the plane in three parts. One on the curve, which satisfies the equality and hence lies on the graph of $y=(x-2)^3$ .

Other portions are to the left and right of the curve. Let us pick two points on either side say $(0,0)$ and $(5,0)$ .

At $(0,0)$ , $y>=(x-2)^3=>0>=-8$ which is true

but at $(5,0)$ , we have $0>=27$ , which is not true.

So, it is only on the left hand side of the curve that points satisfy the inequality $y>=(x-2)^3$

Hence graph appears as follows:

graph{(y-x^3+6x^2-12x+8)>=0 [-10, 10, -5, 5]}

How do you graph the inequality y>=x^3-6x^2+12x-8y>=x^3-6x^2+12x-8?