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Question: Find the domain and the range of a function \[f(x)=\dfrac{\left| x-1 \right|}{x-1}\]...

Find the domain and the range of a function f(x)=x1x1f(x)=\dfrac{\left| x-1 \right|}{x-1}

Explanation

Solution

The domain of the functions is the set of all values that x can take, and the range is the set of all values that f(x)=x1x1f(x)=\dfrac{\left| x-1 \right|}{x-1} can take. Denominator of a fraction cannot be zero. The modulus function x\left| x \right| take the value xx and x-x when x>0x>0 and x<0x<0 respectively.

Complete step-by-step solution:
The given function is f(x)=x1x1f(x)=\dfrac{\left| x-1 \right|}{x-1}
Consider ff be the function defined from the set A to the set B.
Then A is called the domain of the function and contains all possible values x can take.
Also B is called the co-domain of the set.
Then the set of all images of the function, which will be a subset of the co-domain, is called the range of the function.
Now, consider the function given.
f(x)=x1x1f(x)=\dfrac{\left| x-1 \right|}{x-1}
Let's see what all the possible values for x are in order to find the domain.
Division by zero is not specified, as we all know.
As a result, a function's denominator cannot be zero.
x10x-1\ne 0
Adding 1 on both sides, we get:
x1x\ne 1
As a result, the only value that x cannot take is 3.
The domain of this expression is the set of all real numbers except one, which is
\mathbb{R}-\left\\{ 3 \right\\}
Now, the range is the set of all the values taken by f(x)f(x)
We have f(x)=x1x1f(x)=\dfrac{\left| x-1 \right|}{x-1}
Consider x1\left| x-1 \right|
We know that f(x)=xf(x)=\left| x \right| take the value xx and x-x when x>0x>0 and x<0x<0 respectively.
So, we have,
x1=x1\left| x-1 \right|=x-1 If x1>0x-1>0 and x1=(x1)\left| x-1 \right|=-(x-1) if x1<0x-1<0
Further simplifying this we get:
x1=x1\left| x-1 \right|=x-1 If x>1x>1 and x1=(x1)\left| x-1 \right|=-(x-1) if x<1x<1
Case 1:
If x1=x1\left| x-1 \right|=x-1 , then x1x1=x1x1=1\dfrac{\left| x-1 \right|}{x-1}=\dfrac{x-1}{x-1}=1
Case 2:
If x1=(x1)\left| x-1 \right|=-(x-1) then, x1x1=(x1)x1=1\dfrac{\left| x-1 \right|}{x-1}=\dfrac{-\left( x-1 \right)}{x-1}=-1
That is,
f(x)=1f(x)=1 If x>1x>1 and f(x)=1f(x)=-1 If x<1x<1
So f(x)f(x) takes two values 11 and 1-1
This gives the range of the function as \left\\{ -1,1 \right\\} and the domain is R-{1}.

Note: The codomain and domain of a function are also indicated when it is defined. As in this situation, the domain and co-domain do not need to be different. They could also be the same.