Question

How do you find all critical point and determine the min, max and inflection given `f(x)=x^3+x^2-x`?

Answer 1

Critical Points are:
`(-1,1)` min
`(1/3,-5/27)` max

Explanation:

We have `f(x) = x^3 + x^2 - x`

To identify the critical vales, we differentiate and find find values of `x` st `f'(x)=0`

# { ( f'(x) < 0, => f(x) " is decreasing" ), ( f'(x) = 0, => f(x) " is stationary" ), ( f'(x) > 0, => f(x) " is increasing" ) :} #

Differentiating wrt `x`' we have:

`f'(x) = 3x^2 + 2x - 1` .... [1]

At a critical point, `f'(x)=0`

`f'(x)=0 => 3x^2 + 2x - 1 = 0`
`:. (3x-1)(x+1) = 0`
`x=-1,1/3`

Ton find the y-coordinate we substitute the required value into `f(x)`
`x=-1 => f(-1)=-1+1-(-1)=1`
`x=1/3 => f(2/3)=1/27+1/9+1/3=-5/27`

So the critical points are `(-1,1)` and `(1/3,-5/27)`

We identify the nature of these critical points by looking at the sign of second derivative, and

# { ( f''(x) < 0, => f'(x) " is decreasing" => "maximum" ), ( f''(x) = 0, => f'(x) " is stationary" => "inflection" ), ( f''(x) > 0, => f'(x) " is increasing" => "minimum" ) :} #

Differentiating [1] wrt `x` gives s;

`f''(x) = 6x + 2`
`x=-1 => f''(-2)=-6+2 < 0`, ie a maximum
`x=1/3 => f''(1)= 2+2>0`, ie a minimum

Incidental, As this is a cubic with a positive coefficient of `x^3`, we can deduce that the critical point corresponding to the smallest value of `x` must be the maximum and that corresponding to the larger value must be the maximum. It is not possible to have any other possibility!

Answer 2

The max is at `(-1,1)`
The min is at `(1/3,-5/27)`
The inflexion point is at `(-1/3,11/27)`

Explanation:

We have to calculate the first and second derivative.

`f(x)=x^3+x^2-x`

`f'(x)=3x^2+2x-1=(3x-1)(x+1)`

`f'(x)=0` when `x=1/3` and `x=-1`

So, we do a sign chart

`color(white)(aaaa)` `x` `color(white)(aaaa)` `-oo` `color(white)(aaaa)` `-1` `color(white)(aaaa)` `1/3` `color(white)(aaaa)` `+oo`

`color(white)(aaaa)` `f'(x)` `color(white)(aaaaa)` `+` `color(white)(aaaa)` `-` `color(white)(aaaa)` `+`

`color(white)(aaaa)` `f(x)` `color(white)(aaaaaaa)` `uarr` `color(white)(aaaa)` `darr` `color(white)(aaaa)` `uarr`

So, we have a max at `x=-1` and a min at `x=1/3`

To determine the inflexion points, we calculate `f''(x)`

`f''(x)=6x+2`

`f''(x)=0` when `x=-1/3`

The inflexion point is at `x=-1/3`

Also, `f''(-1)=-4<0` which is a max

and `f''(1/3)=4>0` which is a min

graph{x^3+x^2-x [-2.43, 2.436, -1.217, 1.215]}