How do you find the x coordinates of all points of inflection, final all discontinuities, and find the open intervals of concavity for #f(x)=(x^2+1)/(x^2-4)#?

1 Answer
Mar 18, 2017

There are no points of inflection.
The discontinuities are when #x=-2# and #x=2#
#f(x)# is concave up for #x in ]-oo,-2[ uu]2.+oo[#
#f(x)# is concave down for #x in ]-2,2[ #

Explanation:

As we cannot divide by #0#, #x!=-2# and #x!=2#

The domain of #f(x)# is #d_f(x)=RR-{-2,2}#

Therefore, the discontinuities are when #x=-2# and #x=2#

The derivative of a quotient is

#(u/v)'=(u'v-uv')/(v^2)#

We calculate the first derivative of #f(x)#

#u=x^2+1#, #=>#, #u'=2x#

#v=x^2-4#, #=>#, #v'=2x#

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

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

#=(-10x)/(x^2-4)^2#

We can calculate the second derivative of #f(x)#

#u=-10x#, #=>#, #u'=-10#

#v=(x^2-4)^2#, #=>#, #v'=2(x^2-4)*2x=4x(x^2-4)#

#f''(x)=(-10(x^2-4)^2+10x(4x(x^2-4)))/(x^2-4)^4#

#=((x^2-4)(-10x^2+40+40x^2))/(x^2-4)^4#

#=(30x^2+40)/(x^2-4)^3#

#=(10(3x^2+4))/(x^2-4)^3#

The points of inflexion are when, #f''(x)=0#

#f''(x)!=0#, there are no points of inflection.

Let build the chart for concavities

#color(white)(aaaa)##Intervals##color(white)(aaaa)##]-oo,-2[##color(white)(aaaa)##]-2,2[##color(white)(aaaa)##]2,+oo[#

#color(white)(aaaa)##x+2##color(white)(aaaaaaaaaaa)##-##color(white)(aaaaaaaaaaa)##+##color(white)(aaaaaaa)##+#

#color(white)(aaaa)##x-2##color(white)(aaaaaaaaaaa)##-##color(white)(aaaaaaaaaaa)##-##color(white)(aaaaaaa)##+#

#color(white)(aaaa)##Sign f''(x)##color(white)(aaaaaaa)##+##color(white)(aaaaaaaaaaa)##-##color(white)(aaaaaaa)##+#

#color(white)(aaaa)## f(x)##color(white)(aaaaaaaaaaaa)##uu##color(white)(aaaaaaaaaaa)##nn##color(white)(aaaaaaaa)##uu#

Therefore,

#f(x)# is concave up for #x in ]-oo,-2[ uu]2.+oo[#

#f(x)# is concave down for #x in ]-2,2[ #