Question

How do you find the domain and range of \[\sqrt{{{x}^{2}}-4}\]?

Answer 1

Hint: \[\sqrt{{{x}^{2}}-4}\] is the given term. We need to find the domain and the range. For a function, the domain can be derived from the given term. This can be done by finding the roots of the given term. The function value depends upon the resultant roots. The resultant Domain and range can be marked by using graphs.

Complete step by step answer:
The given term is \[\sqrt{{{x}^{2}}-4}\]
Let’s us consider the term \[\sqrt{{{x}^{2}}-4}\] as a function \[f\left( x \right)\]
Now we get the equation as
 \[\Rightarrow f\left( x \right)=\sqrt{{{x}^{2}}-4}\]
Now consider the term under the square root greater than or equal to zero.
Then the equation will be written as
\[\Rightarrow {{x}^{2}}-4\ge 0\]
We can simplify the above equation and we get
\[\Rightarrow \left( x-2 \right)\left( x+2 \right)\ge 0\]
Now let us take
 $ \Rightarrow x-2\ge 0 $
Now the above equation changes as
 $ \Rightarrow x\ge 2 $
 $ \therefore x\ge 2 $
This is one of the values of the domain of \[\sqrt{{{x}^{2}}-4}\]
Now let us take
 $ \Rightarrow x+2\ge 0 $
Now the above equation changes as
 $ \Rightarrow x\ge -2 $
 $ \therefore x\ge -2 $
This is the other value of the domain of \[\sqrt{{{x}^{2}}-4}\]
Therefore the domain of \[f\left( x \right)=\sqrt{{{x}^{2}}-4}\] is \[f\left( x \right)=\left[ -2,2 \right]\]
To find the domain we simplified the term \[\sqrt{{{x}^{2}}-4}\]
Now without simplifying the term we can calculate the domain
Consider the
\[\Rightarrow {{x}^{2}}-4\ge 0\]
Now this equation changes as
\[\Rightarrow {{x}^{2}}\ge 4\]
The exponent of the value $ x $ is $ 2 $ which is as square
Now the square will be the square root of the value $ 4 $
Now it changes as
\[\Rightarrow \left| x \right|\ge 2\]
The mod of $ x $ represented that the term can be negative or positive
\[\Rightarrow -2\ge x\ge 2\]
So the values are $ \left( -2,2 \right) $ .
Hence the domain of \[f\left( x \right)=\sqrt{{{x}^{2}}-4}\]
\[f\left( x \right)=\left[ -2,2 \right]\]
When the term value is \[x=\pm 2\] then the function will be zero
\[f\left( x \right)=0\]
When the term value is \[x=0,f\left( x \right)=2\]this is the maximum value of the function.
The range of \[f\left( x \right)=\sqrt{{{x}^{2}}-4}\]
\[R=\left[ 0,2 \right]\]
\[\therefore x\in \left( -\infty ,-2 \right]\bigcup \left[ 2,\infty \right)\]

Note:
 We solved the range and the domain of \[f\left( x \right)=\sqrt{{{x}^{2}}-4}\] , here we represented the domain and the range in square brackets whereas we represented the interval of the function with both square bracket and parenthesis bracket. The square bracket indicates the endpoint and the parenthesis bracket indicates the point which is not included.