Question

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

Answer 1

I don't actually but I will try my best and I will give you answer it later

Answer 2

Step-by-step explanation:

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

➡️Given real function is f(x) = x/1+x^2.➡️1 + x^2 ≠ 0

➡️Given real function is f(x) = x/1+x^2.➡️1 + x^2 ≠ 0➡️x^2 ≠ -1

➡️Given real function is f(x) = x/1+x^2.➡️1 + x^2 ≠ 0➡️x^2 ≠ -1➡️Domain : x ∈ R

➡️Given real function is f(x) = x/1+x^2.➡️1 + x^2 ≠ 0➡️x^2 ≠ -1➡️Domain : x ∈ R➡️Let f(x) = y

➡️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

➡️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)

➡️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

➡️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 quadratic equation with real roots.

➡️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 quadratic equation with real roots.➡️(-1)^2 – 4(y)(y) ≥ 0

➡️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 quadratic equation with real roots.➡️(-1)^2 – 4(y)(y) ≥ 0➡️1 – 4y^2 ≥ 0

➡️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 quadratic equation with real roots.➡️(-1)^2 – 4(y)(y) ≥ 0➡️1 – 4y^2 ≥ 0➡️⇒ 4y^2 ≤ 1

➡️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 quadratic equation with real roots.➡️(-1)^2 – 4(y)(y) ≥ 0➡️1 – 4y^2 ≥ 0➡️⇒ 4y^2 ≤ 1➡️⇒ y^2 ≤1/4

➡️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 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

➡️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 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

➡️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 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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