Question

the domain of the function f(x)= x/1+x²​

Answer 1

Answer:

Range = [-½, ½]

Step-by-step explanation:

*Given real function is f(x) = x/1+x^2.

➡️1 + x^2 ≠ 0

➡️x^2 ≠ -1

➡️Domain: x ∈ R

➡️Let f(x) = y

➡️y = x/1+x^2

➡️⇒ x = y(1 + x^2)

➡️⇒ yx^2 – x + y = 0

➡️This is a quadratic equation with real roots.

➡️(-1)^2 – 4(y)(y) ≥ 0

➡️1 – 4y^2 ≥ 0

➡️⇒ 4y^2 ≤ 1

➡️⇒ y^2 ≤1/4

➡️⇒ -½ ≤ y ≤ ½

➡️⇒ -1/2 ≤ f(x) ≤ ½

 

➡️Range = [-½, ½]✔

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