How do you solve this problem? The question is attached as an image.

Question

anonymous · Answer

attachment

michele_laino · Answer

hint:
vertical asymptotes are located at x-coordinates for which the denominator is zero

michele_laino · Answer

when 
$${x^2} - 4 = 0\quad ?$$

anonymous · Answer

when x = -2, 2

michele_laino · Answer

ok! So your vertical asymptotes are located at:
x=2
and
x=-2

anonymous · Answer

ok got it

anonymous · Answer

to find what region f(x) is concave you have to find the double derivative and then the zeros right?

michele_laino · Answer

yes! That's right!

michele_laino · Answer

for example: f(x) is concave if f ''(x) <=0

michele_laino · Answer

$$f''\left( x \right) \leqslant 0$$

anonymous · Answer

okay for the second derivative i got (8 x (12+x^2))/(-4+x^2)^3

michele_laino · Answer

please wait, I'm checking your result...

michele_laino · Answer

ok! That's correct!

michele_laino · Answer

now we have to study this inequality:
$$f''\left( x \right) \leqslant 0$$

michele_laino · Answer

$$\frac{{8x\left( {{x^2} + 12} \right)}}{{{{\left( {{x^2} - 4} \right)}^3}}} \leqslant 0$$

anonymous · Answer

sorry i was gone for so long, i kind of had to go somewhere

anonymous · Answer

but for the zeros for that function i got 0 but i'm not sure if that's right

michele_laino · Answer

yes! It is the unique zero of f '' (x)

anonymous · Answer

and the two intervals i have to plug in numbers to are (-inf, 0) and (0, inf)?

michele_laino · Answer

sorry, no I dont' think so!

anonymous · Answer

oh that's why i got the wrong answer. what should the intervals be then?

anonymous · Answer

you have to check both sides of 0.  easy to make a number line with your 0s and then check signs using your factors|dw:1428188422276:dw|

michele_laino · Answer

we have to study the sign of this inequality:
f ''(x)>=0, or its equivalent:
$$\frac{x}{{{x^2} - 4}} \geqslant 0$$

anonymous · Answer

have to include points which give vertical asymptotes

anonymous · Answer

oh ok so (-inf, -2), (-2, 0), (0, 2) and (2, inf) ?

anonymous · Answer

no, not inf.  look at the problem.  the domain is from -15 to 16

anonymous · Answer

and only the intervals in which it is concave up!

anonymous · Answer

ahh yes okay

anonymous · Answer

so it's concave up on the intervals (-2, 0) U (2, 16)

anonymous · Answer

right?

anonymous · Answer

assuming your 2nd derivative is correct (8x(x^2 +12))/(x^2 - 4)

michele_laino · Answer

I think that our function is concave up in this subset:
(-15, -2) union (0,2)

michele_laino · Answer

since the finction f(x) is concave where f ''(x) <0

michele_laino · Answer

oops..function*