Question

Find the stationary points question 11?

Answer 1

Please see below.

Explanation:

.

Stationary points of a function are:

Maximums
Minimums
Points of Inflection

`y=x^3/3-x^2/2-2x-1`

We take the first derivative of the function and set it equal to zero to solve for the roots:

`dy/dx=x^2-x-2=0`

`(x-2)(x+1)=0`

`x=2 and -1`

We plug these into the original function to fins the corresponding `y` values:

`y=8/3-2-4-1=8/3-7=(8-21)/3=-13/3`

`y=-1/3-1/2+2-1=(-2-3+6)/6=1/6`

`(2,-13/3)` is the minimum of the function.

`(-1,1/6)` is the maximum of the function.

Now we take the second derivative of the function and set it equal to zero to find its root:

`(d^2y)/(d^2x)=2x-1=0`

`x=1/2`

`y=1/24-1/8-1-1=(1-3-48)/24=-50/24=-25/12`

We get the point `(1/2,-25/12)`

Now we need to try two values of `x` in the original function, one smaller than `1/2` and one larger. If both corresponding `y`s come out larger than `-25/12`, then this point is another minimum.

If they both come out smaller than `-25/12`, then this point is another maximum.

If one comes out smaller and the other one larger than `-25/12` then this point is the point of inflection.

We will try `-36/12=-3` for the smaller value:

`y=-9-9/2+6-1=-9/2-4=(-9-8)/4=-17/4`

This is smaller than `-25/12`.

We will try `-24/12=-2` as the larger value:

`y=-8/3-2+4-1=-8/3+1=(-8+3)/3=-5/3`

This is larger than `-25/12`

Therefore, `(1/2,-25/12)` is an inflection point:

Here is a graph of the function:

graph{x^3/3-x^2/2-2x-1 [-14.24, 14.24, -7.12, 7.13]}