Functions 1.3 (Domain)

For $f$ to be well defined, the following two restrictions must be true for $x$ :

1. The expression under the radical must be nonnegative. Therefore: $4x^{2}+1\geqslant 0 \Leftrightarrow x\in (-∞,+∞)$
2. The argument of the logarithm must be strictly positive. Therefore: $\sqrt{4x^{2}+1}-2x > 0$

We consider two cases:

Case A: $\boxed{x\geqslant 0}$

$\sqrt{4x^{2}+1}> 2x \Leftrightarrow 4x^{2}+1> 4x^{2}\Leftrightarrow 1> 0$ which is true $\forall x \in [0,+∞)$

Case B: $\boxed{x < 0}$

$\sqrt{4x^{2}+1}> 2x$ . However, we notice that $\sqrt{4x^{2}+1}>0 \enspace$ and $\enspace 2x <0 \enspace \forall x \in (-∞,0)$ .

Therefore, $\sqrt{4x^{2}+1}> 2x \Leftrightarrow \sqrt{4x^{2}+1}-2x > 0$ in both cases.

Therefore, in any case: $x\in (-∞,+∞)$ , which is the domain of the function.