What is the domain and range of #f(x) = (x-1)/(x^2+1)#?

1 Answer
Mar 15, 2018

Answer:

#"Domain":x inRR#
#"Range":f(x)in[-(sqrt(2)+1)/2,(sqrt(2)-1)/2]#

Explanation:

Considering that all real values of #x# will give a non-zero value for #x^2+1#, we can say that for #f(x)#, domain = #x inRR#

For range, we need the maximum and minimum.
#f(x)=(x-1)/(x^2+1)#

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

The maximum and minimum values occur when #f'(x)=0#

#x^2-2x-1=0#

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

#x=(2+-sqrt8)/2=(2+-2sqrt(2))/2=1+-sqrt2#

Now, we input our #x# values into #f(x)#:
#(1+sqrt(2)-1)/((1+sqrt(2))^2+1)=(sqrt(2)-1)/2#
#(1-sqrt(2)-1)/((1-sqrt(2))^2+1)=-(sqrt(2)+1)/2#

#f(x)in[-(sqrt(2)+1)/2,(sqrt(2)-1)/2]#