Question

If f:R→R given by f(x) = {x²-4}/{x²+1}, Identify the type of function.

Answer 1

The function $f:\mathbb{R}\to\mathbb{R}$ is defined as,

$\longrightarrow f(x)=\dfrac{x^2-4}{x^2+1}$

Let,

$\longrightarrow y=\dfrac{x^2-4}{x^2+1}$

$\longrightarrow y(x^2+1)=x^2-4$

$\longrightarrow x^2y+y-x^2+4=0$

$\longrightarrow (y-1)x^2+(y+4)=0$

Now we've a quadratic equation in $x.$

For this equation getting satisfied, there are two conditions.

1. Coefficient of $x^2$ should be non - zero.

$\longrightarrow y-1\neq0$

$\longrightarrow y\neq1$

$\Longrightarrow y\in\mathbb{R}-\{1\}\quad\quad\dots(1)$

2. The discriminant should be non - negative.

$\longrightarrow0^2-4(y-1)(y+4)\geq0$

$\longrightarrow-4(y-1)(y+4)\geq0$

$\longrightarrow(y-1)(y+4)\leq0$

$\Longrightarrow y\in[-4,\ 1]\quad\quad\dots(2)$

Taking intersection of (1) and (2) we get,

$\Longrightarrow y\in[-4,\ 1)$

$\Longrightarrow f(x)\in[-4,\ 1)$

This is the range of our function. But the codomain is $\mathbb{R.}$

Hence $f$ is not an onto function.

We see $f$ is an even function.

$\longrightarrow f(-x)=\dfrac{(-x)^2-4}{(-x)^2+1}$

$\longrightarrow f(-x)=\dfrac{x^2-4}{x^2+1}$

$\longrightarrow f(-x)=f(x)$

This implies, there exists $h\neq0$ such that $f(h)=f(-h).$

Hence $f$ is not a one - one function.

Hence $f$ is not bijective and is not invertible.

Answer 2

$\tt Given \: f(x) = \frac{ {x}^{2} - 4 }{ {x}^{2} + 1}$

$\tt since \: f( - x) = f(x)$

$\tt \implies f(x) = \frac{ {x}^{2} + 1 - 5 }{ {x}^{2} + 1 }$

$\tt \implies 1 - \frac{5}{ {x}^{2} + 1 }$ $\tt f(x)_{m} = - 4$

$\tt f(x)_{M} = 1 \\ \tt Range ∈ (4,1)$ $\tt whereas \: as \: co - domain \: is \: R.$

$\tt As \: range \: is \: not equal \: to \: co - domain, so$

$\tt f(x) \: is \: an \: into \: function.$