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Question: How do you find the x values at which \[f\left( x \right)=\dfrac{\left| x-3 \right|}{x-3}\] is not c...

How do you find the x values at which f(x)=x3x3f\left( x \right)=\dfrac{\left| x-3 \right|}{x-3} is not continuous, which of the discontinuities are removable ?

Explanation

Solution

In order to solve the given question, we can solve by considering the definition of discontinuity. We know that a function is said to be discontinuous at ‘a’ then limxaf(x)\underset{x\to a}{\mathop{\lim }}\,f\left( x \right) is not equal tof(a)f\left( a \right). If the limit does not exist then the function is discontinuous also. In order to find the point of discontinuity of any function, we simply equate the denominator with zero.

Complete step by step answer:
We have given that, f(x)=x3x3f\left( x \right)=\dfrac{\left| x-3 \right|}{x-3}. The given f(x) is continuous at a point x = a in domain if;
limxaf(x)\underset{x\to a}{\mathop{\lim }}\,f\left( x \right) Exists when it is equal to f(a)f\left( a \right).

The given function is discontinuous at x = 3 as it will make the denominator equal to 0. Now,
For x(3,)x\in \left( 3,\infty \right)
(x3)>0, then x3=x3=f(x)\left( x-3 \right)>0,\ then\ \left| x-3 \right|=x-3=f\left( x \right)
Thus,
f(x)=x3x3=x3x3=1f\left( x \right)=\dfrac{\left| x-3 \right|}{x-3}=\dfrac{x-3}{x-3}=1
Now,
For x(,3)x\in \left( -\infty ,3 \right)
(x3)<0, then x3=(x3)=f(x)\left( x-3 \right)<0,\ then\ \left| x-3 \right|=-\left( x-3 \right)=f\left( x \right)
Thus,
f(x)=x3x3=x3x3=1f\left( x \right)=\dfrac{\left| x-3 \right|}{x-3}=-\dfrac{x-3}{x-3}=-1

Therefore, the function f(x)=x3x3f\left( x \right)=\dfrac{\left| x-3 \right|}{x-3} is continuous at \mathbb{R}-\left\\{ 3 \right\\}
Now, taking the limit,
limx3+f(x)=1\underset{x\to {{3}^{+}}}{\mathop{\lim }}\,f\left( x \right)=1
And,
limx3f(x)=1\underset{x\to {{3}^{-}}}{\mathop{\lim }}\,f\left( x \right)=-1
The left limit and the limit from the right are different, thus the discontinuity cannot be removed and the limit given does not exist. Therefore,the limx3f(x)\underset{x\to 3}{\mathop{\lim }}\,f\left( x \right) does not exist.

Hence, the function f(x)=x3x3f\left( x \right)=\dfrac{\left| x-3 \right|}{x-3} has non-removable discontinuity at x = 3.

Note: Discontinuous function is a function that has a discontinuous graph that means the graph which does not have flow of lines of a given function, then the function is said to be a discontinuous function. A continuous function is a function that creates a continuous graph and has a flow of lines, then the given function is said to be a continuous function.