
How do you find the domain of the square root of $1-x^2$?
Answer
451.5k+ views
Hint: The domain of a function is the set of its possible inputs, i.e., the set of input values where for which the function is defined. Here in this we have to find the domain of the function of the square root of 1-x^2. i.e., \[\sqrt {1 - {x^2}} \]. Hence by using the simple arithmetic operations we determine the domain.
Complete step-by-step solution:
Now consider the given function as y. so the function is written as
\[y = \sqrt {1 - {x^2}} \]
The quantity \[1 - {x^2}\] is not a negative value. If it is a negative value then the result will be imaginary. So \[1 - {x^2} \geqslant 0\]--- (1)
add -1 to the equation (1) , the equation (1) is written as
\[ \Rightarrow 1 - {x^2} - 1 \geqslant - 1\]
In the LHS of the above inequality the +1 and -1 will get cancel
\[ \Rightarrow - {x^2} \geqslant - 1\]
Multiply by -1 to the above inequality. On multiplying by -1, the symbol also will change
\[ \Rightarrow {x^2} \leqslant 1\]
Take square root to the above inequality
\[ \Rightarrow \sqrt {{x^2}} \leqslant \sqrt 1 \]
In the LHS of the above inequality the square and square root will get cancel and in the RHS we have both +1 and -1.
\[ \Rightarrow x \leqslant \pm 1\]
Therefore, the domain of x is given as \[[ - 1,1]\]
As a x varies from -1 to 1. The \[{x^2}\] varies from the 0 to 1
Therefore we have
\[0 \leqslant {x^2} \leqslant 1\]
Multiply by -1 to the above inequality
\[ \Rightarrow 0 \geqslant - {x^2} \geqslant - 1\]
add 1 to the above inequality
\[ \Rightarrow 1 \geqslant 1 - {x^2} \geqslant 0\]
Take a principal square root
\[ \Rightarrow 1 \geqslant \sqrt {1 - {x^2}} \geqslant 0\]
Therefore, the range is [0,1]
Note: The set of all possible values which qualify as inputs to a function is known as the domain of the function, or it can also be defined as the entire set of values possible for independent variables. The set of all the outputs of a function is known as the range of the function or after substituting the domain, the entire set of all values possible as outcomes of the dependent variable.
Complete step-by-step solution:
Now consider the given function as y. so the function is written as
\[y = \sqrt {1 - {x^2}} \]
The quantity \[1 - {x^2}\] is not a negative value. If it is a negative value then the result will be imaginary. So \[1 - {x^2} \geqslant 0\]--- (1)
add -1 to the equation (1) , the equation (1) is written as
\[ \Rightarrow 1 - {x^2} - 1 \geqslant - 1\]
In the LHS of the above inequality the +1 and -1 will get cancel
\[ \Rightarrow - {x^2} \geqslant - 1\]
Multiply by -1 to the above inequality. On multiplying by -1, the symbol also will change
\[ \Rightarrow {x^2} \leqslant 1\]
Take square root to the above inequality
\[ \Rightarrow \sqrt {{x^2}} \leqslant \sqrt 1 \]
In the LHS of the above inequality the square and square root will get cancel and in the RHS we have both +1 and -1.
\[ \Rightarrow x \leqslant \pm 1\]
Therefore, the domain of x is given as \[[ - 1,1]\]
As a x varies from -1 to 1. The \[{x^2}\] varies from the 0 to 1
Therefore we have
\[0 \leqslant {x^2} \leqslant 1\]
Multiply by -1 to the above inequality
\[ \Rightarrow 0 \geqslant - {x^2} \geqslant - 1\]
add 1 to the above inequality
\[ \Rightarrow 1 \geqslant 1 - {x^2} \geqslant 0\]
Take a principal square root
\[ \Rightarrow 1 \geqslant \sqrt {1 - {x^2}} \geqslant 0\]
Therefore, the range is [0,1]
Note: The set of all possible values which qualify as inputs to a function is known as the domain of the function, or it can also be defined as the entire set of values possible for independent variables. The set of all the outputs of a function is known as the range of the function or after substituting the domain, the entire set of all values possible as outcomes of the dependent variable.
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