How do you find the range of #f(x)= (x^3+1)^-1#?

1 Answer
Sep 28, 2015

Answer:

#{y|y!=0}#

Explanation:

Do find the range, get the domain of the function's inverse.

#y=(x^3+1)^-1#

Flip #x# and #y#.

#x=(y^3+1)^-1#

Now isolate #y#.

#x=(y^3+1)^-1#
#x=1/(y^3+1)#
#(y^3+1)x=(y^3+1)1/(y^3+1)#
#(y^3+1)x=1#
#y^3x+x=1#
#y^3x+x-x=1-x#
#y^3x=1-x#
#(1/x)y^3x=(1/x)(1-x)#
#y^3=(1-x)/x#
#root(3)(y^3)=root(3)((1-x)/x)#
#y=root(3)((1-x)/x)#
#color(blue)(f^-1(x)=root(3)((1-x)/x))#

Now looking for the domain of the inverse function, we will find that it will only be undefined at #x=0#. So its domain is #{x|x!=0}#.

Therefore, the range of #f(x)=(x^3+1)^-1# is #{y|y!=0}#.