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

Answer:

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

And solving we got:

$x=\pm \sqrt{\frac{72}{12}} =\pm \sqrt{6}$

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

$(-\infty$ Concave up

$x=-\sqrt{6}, y =-180$ inflection point

$(-\sqrt{6}$ Concave down

$x=\sqrt{6}, y=-180$ inflection point

$(\sqrt{6}$ Concave up

Step-by-step 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:

$x=\pm \sqrt{\frac{72}{12}} =\pm \sqrt{6}$

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

$(-\infty$ Concave up

$x=-\sqrt{6}, y =-180$ inflection point

$(-\sqrt{6}$ Concave down

$x=\sqrt{6}, y=-180$ inflection point

$(\sqrt{6}$ Concave up