Question

How do you sketch the graph of `f(x)=x^3-3x^2`?

Answer 1

Calculate derivative :

`f'(x) = 3 x^2 - 6 x = 3x (x-2)`

So you can study the sign of `f'(x)` :

• `f'(x) < 0 iff x in ]0,2[`
• `f'(x) > 0 iff x in ]-oo, 0[ cup ]2,+oo[`.

You get now the variations of `f` :
- `f` is decreasing on `[0,2]`
- `f` is creasing on `[-oo, 0]` and on `[2,+oo]`.

Remark that `f` has :
- local maximum in 0 with `f(0)=0`
- local minimum in 2 with `f(2) =-4`.
There are two horizontal tangents at 0 and at 2.

• For the limits, apply the rule :
  `lim_{x->-oo} x^3-3x^2 = lim_{x->-oo} x^3 = -oo`
  `lim_{x->+oo} x^3-3x^2 = lim_{x->+oo} x^3 = +oo`

Finally :

graph{x^3 - 3x^2 [-8.42, 13.78, -6.62, 4.48]}

Answer 2

You can start by setting `x=0` that gives you `y=f(0)=0` so your curve passes through the origin.
Setting `y=0` you get `x^3-3x^2=0` that gives `x=0` and `x=3`
When `x->+oo` `f(x)->+oo` as well while when `x->-oo` then `f(x)->-oo`.
Points of maximum or minimum are found by setting the first derivative equal to zero:
`f'(x)=3x^2-6x`
and
`3x^2-6x=0` gives: `3x(x-2)=0`
and `x=0` and `x=2`
When `x=0` `y=0` and when `x=2` `y=-4`
Setting the second derivative equal to zero will give us inflection point(s):
`f''(x)=6x-6`
and
`6x-6=0`
so that `x=1` and `y=-2`
And finally: