Question

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

Answer 1

Step-by-step explanation:

Hey user;

firstly,

$1 + {x}^{2} \neq 0 \\ \\ {x}^{2} \neq -1$

So domain is $R$

For finding range;

Just assume one variable as 'y'

y = f(x)

$y = \frac{{x}^{2}}{1 + {x}^{2}} \\ \\ y = {x}^{2}(y - 1 ) \\ \\ x = \pm \sqrt{ \frac{y}{y-1}} \\ \\ y - 1 \neq 0 \\ \\ y \neq -1 \\ \\ \therefore \: Range \: is \leadsto R - (-1)$

Answer 2

Answer:

Domain is { R }

Range is [ 0 , - 1 )

Step-by-step-explanation:

Function is defined as a relation between two sets. Let these sets,here, be A and B. According to the definition for functions, there will be an image( from B ) for a each and every unique a € A { a belonging to A }.

Real valued functions are the same, between two sets of real numbers, which says for every real number belonging to set A ( belonging to R or its subset ) there will be a real number( belonging to R ) from set B.

Since we deal in real numbers, domain as well as range will be taken from R( set of real numbers ) or a subset of R.

Therefore, value of the given should be a real number. For this, denominator should not be 0.

Here,

we find the the denominator is 1 + x^2, we know square of any real number will always be a positive real number. Therefore, x^2 + 1 will always be a positive real number ( not 0 ).

Hence,

Domain of the given function is { R }.

Let x^2 / ( 1 + x^2 )^2 = y

= > x^2 = y + yx^2

= > x^2 - yx^2 = y

= > x^2( 1 - y ) = y

= > x^2 = y / ( 1 - y )

= > x = ±√{ y / ( 1 - y ) }

For a positive real number, 1 - y will always be a positive real number.

1 - y > 0

1 > y

Since y is a positive real number { y = x^2 / ( 1 + x^2 ) }, its value can't be negative. So, 1 > y ≥ 0

Hence,

Range is [ 0 , - 1 ).