Question

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

Answer 1

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$\Large\fbox{\color{purple}{QUESTION}}$

Find the domain and range of the real function f(x) = x/1+x^2.

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$\bf\Huge\red{\mid{\overline{\underline{ ANSWER }}}\mid }$

➡️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]

$\Large\mathcal\green{FOLLOW \: ME}$

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Answer 2

Answer:

SOLUTION:-

➡️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]✔

Step-by-step explanation:

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