Answer
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Hint: The domain of a function is the complete step of possible values of the independent variable. That is the domain is the set of all possible ‘x’ values which will make the function ‘work’ and will give the output of ‘y’ as a real number. The range of a function is the complete set of all possible resulting values of the dependent variable, after we have substituted the domain.
Complete step by step solution:
The domain of a function is the complete set of possible values of the independent variable. The domain of the expression is all real numbers except where the expression is undefined.
Here independent variable is x
Given, \[f\left( x \right) = \dfrac{{3x + 1}}{{x - 5}}\].
Now we have to consider the values of x. Here x can take all the numbers from the set of real numbers. But it can’t take the number 5.
If we take the value of x as 5, then the function will be
\[ \Rightarrow f\left( 5 \right) = \dfrac{{3(5) + 1}}{{5 - 5}}\]
The denominator of the function will be zero and the function will be undefined.
Therefore the domain of the function is \[R - \{ 5\} \]
This represents the x can take all the values from the set of real numbers but not the number 5.
The domain can also be represented as
\[( - \infty ,5] \cup [5,\infty )\]
The ( ) represents the open set where the limit points are considered. The [ ] represents the closed set where the limit points are not considered.
So, the correct answer is “\[( - \infty ,5] \cup [5,\infty )\]”.
Note: The domain where the x values ranges and the range where the y values ranges. Since the equation is a quadratic equation we find the value of x and then we determine the value of y for the values of x. The students must know the simple arithmetic operations.
Complete step by step solution:
The domain of a function is the complete set of possible values of the independent variable. The domain of the expression is all real numbers except where the expression is undefined.
Here independent variable is x
Given, \[f\left( x \right) = \dfrac{{3x + 1}}{{x - 5}}\].
Now we have to consider the values of x. Here x can take all the numbers from the set of real numbers. But it can’t take the number 5.
If we take the value of x as 5, then the function will be
\[ \Rightarrow f\left( 5 \right) = \dfrac{{3(5) + 1}}{{5 - 5}}\]
The denominator of the function will be zero and the function will be undefined.
Therefore the domain of the function is \[R - \{ 5\} \]
This represents the x can take all the values from the set of real numbers but not the number 5.
The domain can also be represented as
\[( - \infty ,5] \cup [5,\infty )\]
The ( ) represents the open set where the limit points are considered. The [ ] represents the closed set where the limit points are not considered.
So, the correct answer is “\[( - \infty ,5] \cup [5,\infty )\]”.
Note: The domain where the x values ranges and the range where the y values ranges. Since the equation is a quadratic equation we find the value of x and then we determine the value of y for the values of x. The students must know the simple arithmetic operations.
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