Respuesta :
Answer:
[tex] y'' =12x^2 -72=0[/tex]
And solving we got:
[tex] x=\pm \sqrt{\frac{72}{12}} =\pm \sqrt{6}[/tex]
We can find the sings of the second derivate on the following intervals:
[tex] (-\infty<x< -\sqrt{6}) , y'' = +[/tex] Concave up
[tex]x=-\sqrt{6}, y =-180[/tex] inflection point
[tex] (-\sqrt{6} <x< \sqrt{6}), y'' =-[/tex] Concave down
[tex]x=\sqrt{6}, y=-180[/tex] inflection point
[tex] (\sqrt{6}<x< \infty) , y'' = +[/tex] Concave up
Step-by-step explanation:
For this case we have the following function:
[tex] y= x^4 -36x^2[/tex]
We can find the first derivate and we got:
[tex] y' = 4x^3 -72x[/tex]
In order to find the concavity we can find the second derivate and we got:
[tex] y'' = 12x^2 -72[/tex]
We can set up this derivate equal to 0 and we got:
[tex] y'' =12x^2 -72=0[/tex]
And solving we got:
[tex] x=\pm \sqrt{\frac{72}{12}} =\pm \sqrt{6}[/tex]
We can find the sings of the second derivate on the following intervals:
[tex] (-\infty<x< -\sqrt{6}) , y'' = +[/tex] Concave up
[tex]x=-\sqrt{6}, y =-180[/tex] inflection point
[tex] (-\sqrt{6} <x< \sqrt{6}), y'' =-[/tex] Concave down
[tex]x=\sqrt{6}, y=-180[/tex] inflection point
[tex] (\sqrt{6}<x< \infty) , y'' = +[/tex] Concave up
