Question

find the domain and range of the following 1. f(x)=1/x-1 2.f(x)=√x+1

Answer 1

I hope it will be answer of your question

Answer 2

Answer:

$1. \:\:f(x) = \dfrac{1}{x-1}$

We know that denominator cannot be 0. Hence x cannot be 1 since denominator will become 0 if x is 1. So, x can take all real values except 1.

Hence the domain of x can be written as: {R} - 1, where {R] stand for all real values.

Now since the denominator is having the x, the function cannot take a greater value than 1, since numerator remains the same. So there are 2 cases.

Case 1: When x < 0

Then Denominator has a negative value and hence the value will be greater than -1.

Case 2: When x > 0

Then the denominator will have a positive value and is less than 1.

So the possible range of values lies between 1 and -1.

Hence Range of the given f(x) is ( -1, 1 )  

[ Open bracket, since it cannot be 1 and -1. }

=======================================================

$2. \;\: f(x) = \sqrt{x} + 1$

We know that. inside a square root, negative numbers cannot be present. Hence x must take only positive values.

Hence x belongs to positive values only.

Hence Domain of f(x) = All positive real values ( or ) [ 0,∞ ]

Since the Root value is always going to be positive, and 1 added to it will always be positive, the range is all real values starting from 1.

Range of f(x) = ( 0,∞ )

This is the required answer !!

Hope it helped !!