Question

Show that for all Real x, sqrt(x^2 + 1) + x > 0 ( using disjunction of cases)

Answer 1

- first of all you need to know where is defined the square root function

\[f(x) = \sqrt{x} \]

Answer 2

Its Domain is the Non-Negative Real Numbers: [0, +∞).

Answer 3

really

where does the function \(\sqrt{x}\) feature in this question?