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Question: Find the domain and range of a function \(f(x) = \dfrac{{\left| {x - 3} \right|}}{{x - 3}}\). A) \...

Find the domain and range of a function f(x)=x3x3f(x) = \dfrac{{\left| {x - 3} \right|}}{{x - 3}}.
A) R,1,1\mathbb{R},\\{ - 1,1\\}
B) R3,1,1\mathbb{R} - \\{ 3\\} ,\\{ - 1,1\\}
C) RT,R{\mathbb{R}^{\text{T}}},\mathbb{R}
D) None of these

Explanation

Solution

Domain of the function is the set of all values taken by xx and range is the set of all values taken by f(x)f(x). Denominator of a fraction cannot be zero. The modulus function x\left| x \right| takes the value xx and x - x when x>0x > 0 and x<0x < 0 respectively.

Useful formula:
The function f(x)=xf(x) = \left| x \right| takes the value xx if x>0x > 0 and x - x if x<0x < 0.

Complete step by step solution:
The given function is f(x)=x3x3f(x) = \dfrac{{\left| {x - 3} \right|}}{{x - 3}}.
Let ff be a function defined from the set AA to the set BB.
Then AA is called the domain of the function and contains all possible values xx can take.
Also BB 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)=x3x3f(x) = \dfrac{{\left| {x - 3} \right|}}{{x - 3}}
To find the domain let us check what all values xx can take here.
We know that division by zero is not defined.
So the denominator of a function cannot be zero.
This gives,
x30x - 3 \ne 0
Adding 33 on both sides we get,
x3x \ne 3
So the only value which could not be taken by xx is 33.
This gives the domain is the set of all real numbers except three, that is R3\mathbb{R} - \\{ 3\\} .
Now the range is the set of all values taken by f(x)f(x).
We have f(x)=x3x3f(x) = \dfrac{{\left| {x - 3} \right|}}{{x - 3}}
Consider x3\left| {x - 3} \right|.
We know that f(x)=xf(x) = \left| x \right| takes the value xx if x>0x > 0 and x - x if x<0x < 0.
So we have,
x3=x3\left| {x - 3} \right| = x - 3 if x3>0x - 3 > 0 and x3=(x3)\left| {x - 3} \right| = - (x - 3) if x3<0x - 3 < 0
x3=x3\left| {x - 3} \right| = x - 3 if x>3x > 3 and x3=(x3)\left| {x - 3} \right| = - (x - 3) if x<3x < 3
If x3=x3\left| {x - 3} \right| = x - 3, then x3x3=x3x3=1\dfrac{{\left| {x - 3} \right|}}{{x - 3}} = \dfrac{{x - 3}}{{x - 3}} = 1
And if x3=(x3)\left| {x - 3} \right| = - (x - 3), then x3x3=(x3)x3=1\dfrac{{\left| {x - 3} \right|}}{{x - 3}} = \dfrac{{ - (x - 3)}}{{x - 3}} = - 1
That is,
f(x)=1f(x) = 1 if x>3x > 3 and f(x)=1f(x) = - 1 if x<3x < 3.
So f(x)f(x) takes two values 11 and 1 - 1.
This gives the range of the function is 1,1\\{ - 1,1\\} .

Therefore the answer is option B.

Note:
When a function is defined, its domain and co-domain are also mentioned. The domain and co-domain need not be different as in this case. They may be the same as well.