Question

find domain and range of function f(x)=×/1+Xsquare

Answer 1

f(x) = x / 1 + x²      1+x² ≠ 0   x² ≠ -1 ⇒ x² can never be -1

so f(x) is defined for all valus of x

hence  domain = x ∈ R

now   f(x) = y  = x/1 + x² ⇒  y + yx² = x ⇒  yx² - x + y = 0
⇒  forms a quadratic eq whose roots should be real

⇒ D ≥ 0 ⇒ b² - 4ac ≥ 0  ⇒  1 - 4y² ≥ 0  ⇒  (1 - 2y)( 1+ 2y) ≥ 0

⇒   y ∈ ( - ∞ . -1/2] U [ 1/2 , ∞) is the range of function

hope u'll get it ...........if any query, then ask.