Question

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

Answer 1

Answer:

For all x ∈ R , the function f(x) exists.

Therefore, R is the domain of f(x).

Answer 2

$\huge \bold{ \underline { \bf{Question \to}}}$

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

$\huge \bold{ \underline { \bf{Solution \to}}}$

Given,

real function is $\sf { f(x) = x/1+x^2}$

$\sf {⇒1 + x² ≠ 0}$

$\sf {⇒x² ≠ -1}$

$\Large\tt {\red{Domain : x ∈ R}}$

Let $\sf {f(x) = y}$

$\sf {y = \frac{x}{1}+x²}$

$\sf {⇒ x = y(1 + x²)}$

$\sf {⇒ yx^2 \:– \: x + y = 0}$

This is quadratic equation with real roots ;

$\sf {⇒(-1)^2 – 4(y)(y) ≥ 0}$

$\sf {⇒1 – 4y^2 ≥ 0}$

$\sf {⇒ 4y^2 ≤ 1}$

$\sf {⇒ y^2 ≤1/4}$

$\sf {⇒ -½ ≤ y ≤ ½}$

$\sf {⇒ -½ ≤ f(x) ≤ ½}$

$\Large\tt {\red{Range \implies [-½, ½]}}$