Question

find the domain and range of the real function f(x)=x/1+x²​

Answer 1

Answer:

domain =x belongs to real no and range is y belongs to (0, 1)

Step-by-step explanation:

Answer 2

Given real function is f(x) = x/1+x²

1 + x² ≠ 0

x² ≠ -1

Domain : x ∈ R

Let f(x) = y

y = x/1+x²

⇒ x = y(1 + x2)

⇒ yx² – x + y = 0

This is quadratic equation with real roots.

(-1)² – 4(y)(y) ≥ 0

1 – 4y² ≥ 0

⇒ 4y² ≤ 1

⇒ y² ≤1/4

⇒ -½ ≤ y ≤ ½

⇒ -1/2 ≤ f(x) ≤ ½

Range = [-½, ½]