Find the domain of $\sqrt {x + 7} $?

Hint: For the domain we have to find the value of $x$ for which the given function is always defined , in this case we will put the value under root greater than equal to zero, and then find the values of $x$ which it is defined.

Complete step-by-step solution:
Given: Let the function be, $f(x) = \sqrt {x + 7} $
To find: We have to find the domain of the given function $f(x) = \sqrt {x + 7} $
Given real function is $f(x) = \sqrt {x + 7} $
As we know if the value of the square root is negative then the function is not defined.
So the value of the square root is always greater than or equal to zero.
Thus $f(x) = \sqrt {x + 7} \geqslant 0$………………….$(1)$
Now squaring both sides, we have
$(x + 7) \geqslant 0$
On rewriting we get,
$x \geqslant - 7$
Therefore the domain of $f$ is the set of all the numbers which are greater than or equal to $ - 7$
That is the domain of the $f(x)$ is $[ - 7,\infty )$ (Where domain is greater than or equal to $ - 7$ so at $ - 7$ there is a closed interval and at infinity there is an open interval).
Now from the equation $(1)$
$\sqrt {x + 7} \geqslant 0$
But $f(x) = \sqrt {x + 7} $
$ \Rightarrow f(x) \geqslant 0$
Therefore the domain of $f$ is the set of real numbers greater than or equal to $0$ (zero).
That is the domain of $f = [ - 7,\infty )$(domain is greater than or equal to $0$ (zero) so at $ - 7$ there is a closed interval and at infinity there is an open interval).
Hence, this is the required answer.

Note: Domain is an independent set of those values for a given function which on substitution always gives values of result.
Solutions of inequalities are always written in the form of intervals like close and open intervals.