Question
Find the domain and the range of the real function: $f(x)=\frac{x^2-16}{x-4}$

Answer

Here we are given that, $f(x)=\frac{x^2-16}{x-4}$
Need to find: where the function is defined. 
Let, $f(x)=\frac{x^2-16}{x-4}=y$
To find the domain of the function f(x) we need to equate the denominator of the function to 0 
Therefore, 
x - 4 = 0 or x = 4 
It means that the denominator is zero when $x=4$
So, the domain of the function is the set of all the real numbers except 4
The domain of the function, $D _{\{ f ( x )\}}=(-\infty, 4) \cup(4, \infty)$
Now if we put any value of $x$ from the domain set the output value will be either (-ve) or (+ve), but the value will never be 8
So, the range of the function is the set of all the real numbers except 8
The range of the function, $R _{ f ( x )}=(-\infty, 8) \cup(8, \infty)$

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