Question

f: R—>(-1,1),f(x) = x/1+| x | , Is the given function one-one or onto?

Answer 1

f : R -----> (-1, 1) , f(x) = x/(1 + |x|)

case 1 :- if x ≥ 0 , f(x) = x/(1 + x)
now, taking two different points x1 and x2 from. domain of f(x) in such a way that f(x1) = f(x2).
f(x1) = f(x2)
x1/(1 + x1) = x2/(1 + x2)
x1(1 + x2) = x2(1 + x1)
x1 + x1x2 = x2 + x1x2
x1 = x2
so, f is one - one function.

case 2 :- x < 0 , f(x) = x/(1 - x)
taking two different points x1 and x2 from the domain of function , f(x) in such a way that
f(x1) = f(x2)
x1/(1 - x1) = x2/(1 - x2)
x1(1 - x2) = x2(1 - x1)
x1 - x1x2 = x2 - x1x2
x1 = x2
so, f is one - one function.

hence, given function function is one - one function.

now checking function is onto or not.
case 1 :- x ≥ 0, f(x) = y = x/(1 + x) ,
y + xy = x
x = y/(1 - y) , x ≥ 0
f(y) = y/(1 - y), x ≥ 0

case 2 :- x < 0 , f(x) = y = x/(1 -x)
y - xy = x
x = y/(1 + y)
f(y) = y/(1 + y) , x < 0
$here \: \: \: y\in\{x\in R:-1 < x < 1 \}$
hence, domain of f(y) belongs to -1 < x < 1
hence, range of f(x) belongs to -1 < x < 1
here we see, co-domain = range
so, f is onto function.

finally we get, f is one - one and onto