Question

How do you find the domain and range of \[y=\sqrt{{{x}^{2}}-9}\] ?

Answer 1

Hint: We are given a square root function, to find the value of \[x\] such that the given function accepts the input value. For this the part inside the square root must be positive for then the function will be defined. So, put \[{{x}^{2}}-9\] greater than zero which implies \[{{x}^{2}}\] must be greater than \[9\]. Now using fundamentals of mathematics, we will get \[x\] belongs to \[(-\infty ,-3]\cup [3,\infty )\] means on putting the value of \[x\] from this region the function will be defined otherwise not. Now the range varies according to the domain here the function is square root means it gives minimum value zero and by putting the \[x\] very large in positive and negative both we would get a large positive value.

Complete step by step solution:
Given that, \[y=\sqrt{{{x}^{2}}-9}\]
We have to find the domain of function,
To make the define, the input inside the square root must be non-negative for this function
\[\Rightarrow {{x}^{2}}-9\ge 0\]
\[\Rightarrow (x-3)(x+3)\ge 0\]
Since the product of these linear terms is greater than zero, means either both terms must be positive or both must be negative
\[\Rightarrow (x-3)\ge 0\] and \[(x+3)\ge 0\] Or \[(x-3)\le 0\] and \[(x+3)\le 0\]
Now first case: \[(x-3)\ge 0\] and \[(x+3)\ge 0\]
\[\Rightarrow x\ge 3\] and \[x\ge -3\]
Now these both conditions must satisfy simultaneously,
Taking common region between them
\[\Rightarrow x\ge 3\]
\[\Rightarrow x\in [3,\infty )\]
Now second case: \[(x-3)\le 0\] and \[(x+3)\le 0\]
\[\Rightarrow x\le 3\] and \[x\le -3\]
These both must satisfy
Similarly,
\[\Rightarrow x\le -3\]
\[\Rightarrow x\in [-3,\infty )\]
Since from the above first and second case either of the conditions can be true, so we will take union of these.
\[\Rightarrow x\in [-3,\infty )\cup [3,\infty )\]
This is the domain of the given function
Now we are calculating the range of this given function
Since the range of function depends on its domain, means on putting domain as input value we will get the output value.
We have \[y=\sqrt{{{x}^{2}}-9}\]
Since the domain is \[x\in [-3,\infty )\cup [3,\infty )\]
And in this function, we have variable \[{{x}^{2}}\] means it is symmetric with \[x\] means it is independent of sign.
Now on putting the least value in \[x\], we get zero as our output and putting a very large number closer to infinity we will get a large number closer to infinity.
Thus, the range of this function is from zero to infinity.
\[\Rightarrow R\in [0,\infty )\]

Hence domain of function is \[x\in [-3,\infty )\cup [3,\infty )\] and range is \[R\in [0,\infty )\].

Note:
When we have to calculate the domain of a function just find the region for \[x\] that we can put in the function such that function must be defined for that particular region. And on putting that we will calculate the range. Note that while writing the region the bracket we use near the infinity is a small bracket as we don’t know the exact value of infinity otherwise the block brackets include the infinity part which we have not known.