How do you find the range of #f(x)= x^2/(1-x^2)#?

Question

29937 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2018-03-21T21:53:14

#(-oo, -1) uu [0, oo)#

Given:

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

Let #y = f(x)# and attempt to solve for #x#...

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

Add #1# to both ends to get:

#y + 1 = 1/(1-x^2)#

Multiply both sides by #(1-x^2)/(y+1)# to get:

#1-x^2 = 1/(y+1)#

Add #x^2-1/(y+1)# to both sides to get:

#1-1/(y+1) = x^2#

In order for this to have a real valued solution, we require:

#1-1/(y+1) >= 0#

That is:

#y/(y+1) >= 0#

Hence we require one of:

#(y >= 0 ^^ y+1 > 0) rarr y in [0, oo)#

#(y < 0 ^^ y+1 < 0) rarr y in (-oo, -1)#

So the range of #f(x)# is #(-oo, -1) uu [0, oo)#

graph{x^2/(1-x^2) [-10, 10, -5, 5]}