The function is

#f(x)=4/(x^2-1)#

The domain of #f(x)# is #x in (-oo, -1)uu(-1, 1)uu(1, +oo)#

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)##uu##color(white)(aaaaaaaaaa)##nn##color(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)#