Question

let { f = (x, x²/1+x²) :x €R } be a function from R to R. Determine the range of f.​

Answer 1

$\large\underline{\bold{Given - }}$

$\sf \: Let \: f = \bigg(x,\dfrac{ {x}^{2} }{1 + {x}^{2} } \bigg), \: x \in \: R \: be \: a \: function \: from \: R \to \: R$

$\large\underline{\bold{To\:Find - }}$

$\sf \: Range \: of \: f$

$\large\underline{\bold{Concept \: Used - }}$

The steps for algebraically finding the range of a function are:

• Write down y = f(x)

• solve the equation for x, giving something of the form x = g(y).

• Find the domain of g(y), and this will be the range of f(x).

Let's do now!!

$\large\underline{\bold{Calculation}}$

• Given that

$\rm :\longmapsto\:f = \bigg(x,\dfrac{ {x}^{2} }{1 + {x}^{2} } \bigg)$

• To find range of f,

$\rm :\longmapsto\:Let \: y \: = \dfrac{ {x}^{2} }{1 + {x}^{2} }$

$\rm :\longmapsto\:y + y {x}^{2} = {x}^{2}$

$\rm :\longmapsto\:y = {x}^{2} - y {x}^{2}$

$\rm :\longmapsto\:y = {x}^{2} (1 - y)$

$\rm :\longmapsto\: {x}^{2} = \dfrac{y}{1 - y}$

$\rm :\longmapsto\:x = \sqrt{\dfrac{y}{1 - y} }$

So,

$\rm :\implies\:x \: is \: defined \: iff \: y \geqslant 0 \: and \: 1 - y > 0$

$\rm :\implies\:y \geqslant 0 \: and \: y < 1$

$\rm :\implies\: \boxed{ \bf \: y \in \: [0,1)}$

Hence,

$\rm :\longmapsto\:Range \: of \: f = \bigg(x,\dfrac{ {x}^{2} }{1 + {x}^{2} } \bigg) \: is \: [0,1)$