
How do you find the domain and range of \[x + \sqrt {x - 1} \] ?
Answer
490.2k+ views
Hint: Let the given function be \[f\left( x \right)\] .We need to find the domain and range of the given function \[f\left( x \right)\] . In order to solve this, we will use the basic concept of domain and range. The domain is the set of all possible values of \[x\] for which the function \[f\left( x \right)\] will be defined and the range is the possible range of values that the function \[f\left( x \right)\] can attain for those values of \[x\] which are in the domain of \[f\left( x \right)\].
Complete step by step answer:
Let the given function be \[f\left( x \right)\]
Therefore, \[f\left( x \right) = x + \sqrt {x - 1} {\text{ }} - - - \left( i \right)\]
As we know if the value of the square root is negative, then the function is not defined.
Therefore, the value of square root is always greater than or equal to zero.
\[ \Rightarrow \sqrt {x - 1} \geqslant 0\]
On squaring both sides, we have
\[ \Rightarrow \left( {x - 1} \right) \geqslant 0\]
On adding \[1\] both sides, we get
\[ \Rightarrow x \geqslant 1\]
Therefore, for \[x \geqslant 1\] the function \[f\left( x \right)\] is defined.
And we know that the domain is the set of all possible values of \[x\] for which the function \[f\left( x \right)\] is defined.
Therefore, the domain of \[f\] is the set of all real numbers greater than or equal to \[1\].Hence, the domain of the function \[x + \sqrt {x - 1} \] is \[\left[ {1,\infty } \right)\].Now, we know that the range is the possible range of values that the function \[f\left( x \right)\] can attain for those values of \[x\] which are in the domain of \[f\left( x \right)\].So, when \[x = 1\]. The range will be \[1 + \sqrt {1 - 1} = 1\]. Now for \[x \geqslant 1\] , the range of \[f\left( x \right)\] is \[f\left( x \right) \geqslant 1\]. Therefore, the range of \[f\] is the set of all real numbers greater than or equal to \[1\].
Hence, the range of the function \[x + \sqrt {x - 1} \] is \[\left[ {1,\infty } \right)\].
Note: Whenever we face such types of problems, the key concept is to be clear about the definitions of domain and range. Also, remember that the infinity symbol is always accompanied by a round bracket. While answering if you write \[\left[ {1,\infty } \right]\] then this would be the wrong answer. So, while writing in the notation form, take care of brackets.
Complete step by step answer:
Let the given function be \[f\left( x \right)\]
Therefore, \[f\left( x \right) = x + \sqrt {x - 1} {\text{ }} - - - \left( i \right)\]
As we know if the value of the square root is negative, then the function is not defined.
Therefore, the value of square root is always greater than or equal to zero.
\[ \Rightarrow \sqrt {x - 1} \geqslant 0\]
On squaring both sides, we have
\[ \Rightarrow \left( {x - 1} \right) \geqslant 0\]
On adding \[1\] both sides, we get
\[ \Rightarrow x \geqslant 1\]
Therefore, for \[x \geqslant 1\] the function \[f\left( x \right)\] is defined.
And we know that the domain is the set of all possible values of \[x\] for which the function \[f\left( x \right)\] is defined.
Therefore, the domain of \[f\] is the set of all real numbers greater than or equal to \[1\].Hence, the domain of the function \[x + \sqrt {x - 1} \] is \[\left[ {1,\infty } \right)\].Now, we know that the range is the possible range of values that the function \[f\left( x \right)\] can attain for those values of \[x\] which are in the domain of \[f\left( x \right)\].So, when \[x = 1\]. The range will be \[1 + \sqrt {1 - 1} = 1\]. Now for \[x \geqslant 1\] , the range of \[f\left( x \right)\] is \[f\left( x \right) \geqslant 1\]. Therefore, the range of \[f\] is the set of all real numbers greater than or equal to \[1\].
Hence, the range of the function \[x + \sqrt {x - 1} \] is \[\left[ {1,\infty } \right)\].
Note: Whenever we face such types of problems, the key concept is to be clear about the definitions of domain and range. Also, remember that the infinity symbol is always accompanied by a round bracket. While answering if you write \[\left[ {1,\infty } \right]\] then this would be the wrong answer. So, while writing in the notation form, take care of brackets.
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