The function is
f(x)=4/(x^2-1)f(x)=4x2−1
The domain of f(x)f(x) is x in (-oo, -1)uu(-1, 1)uu(1, +oo)x∈(−∞,−1)∪(−1,1)∪(1,+∞)
Calulate the first derivative with the quotient rule
(u/v)'=(u'v-uv')/(v^2)
u=4, =>, u'=0
v=x^2-1, =>, v'=2x
Therefore,
f'(x)=(0*(x^2-1)-4*2x)/(x^2-1)^2=-(8x)/(x^2-1)^2
f'(x)=0, =>, x=0
There is a critical point at (0, -4)
Calulate the second derivative with the quotient rule
u=-8x, =>, u'=-8
v=(x^2-1)^2, =>, v'=4x(x^2-1)
f''(x)=(-8(x^2-1)^2+32x^2(x^2-1))/(x^2-1)^4
=(-8x^2+8+32x^2)/(x^2-1)^3
=(24x^2+8)/(x^2-1)^3
Therefore,
f''(x)!=0, AA x in "domain"
Build a variation chart to determine the concavities
color(white)(aaaa)"Interval"color(white)(aaaa)(-oo,-1)color(white)(aaaa)(-1,1)color(white)(aaaa)(1,+oo)
color(white)(aaaa)"Sign f''(x)"color(white)(aaaaaaa)+color(white)(aaaaaaaaaa)-color(white)(aaaaaaaa)+
color(white)(aaaa)" f(x)"color(white)(aaaaaaaaaaa)uucolor(white)(aaaaaaaaaa)nncolor(white)(aaaaaaaa)uu
Finally,
f(x) is convex for x in (-oo,-1)uu(1, +oo)
f(x) is concave for x in (-1,1)
