Question

please show on graph (with x and y coordinates) state where the function x^4-36x^2 is non-negative, increasing, concave up​

Answer 1

`y'' =12x^2 -72=0`

And solving we got:

`x=\pm \sqrt{(72)/(12)} =\pm √(6)`

We can find the sings of the second derivate on the following intervals:

`(-\infty<x< -√(6)) , y'' = +` Concave up

`x=-√(6), y =-180` inflection point

`(-√(6) <x< √(6)), y'' =-` Concave down

`x=√(6), y=-180` inflection point

`(√(6)<x< \infty) , y'' = +` Concave up

Explanation:

For this case we have the following function:

`y= x^4 -36x^2`

We can find the first derivate and we got:

`y' = 4x^3 -72x`

In order to find the concavity we can find the second derivate and we got:

`y'' = 12x^2 -72`

We can set up this derivate equal to 0 and we got:

`y'' =12x^2 -72=0`

And solving we got:

`x=\pm \sqrt{(72)/(12)} =\pm √(6)`

We can find the sings of the second derivate on the following intervals:

`(-\infty<x< -√(6)) , y'' = +` Concave up

`x=-√(6), y =-180` inflection point

`(-√(6) <x< √(6)), y'' =-` Concave down

`x=√(6), y=-180` inflection point

`(√(6)<x< \infty) , y'' = +` Concave up