calculus - What is the domain of the function $f(x)$ and what is the value of the parameter $a$ for which the function is always positive?

I have the following function:

$f: D \rightarrow \mathbb{R},$ $f(x)= \dfrac{ln(x+a)}{\sqrt{x}}$

$a \in \mathbb{R}$

And I am asked $2$ things:

• Find the domain $D$ of the function $f(x)$.


• Find the values of $a$ such that $f(x) > 0, \forall x \in
  \mathbb{R}$.

Concerning point $1$, I applied the following conditions:

$$x > 0 \Rightarrow x \in (0, + \infty)$$

$$x+a>0 \Rightarrow x > -a \Rightarrow x \in (-a, + \infty)$$

Combining these $2$ conditions, I got:

$$x \in (max(-a, 0), + \infty)$$

However, when I checked my textbook, I found that the answer they have listed is this:

$$x \in \bigg ( \dfrac{-a+ |a| }{2}, + \infty \bigg )$$

Can somebody explain why are these $2$ answers equivalent (that is, if I didn't do any mistakes and they are indeed equivalent)?

And concerning the $2 ^ {nd}$ point, I tried to find the derivative thinking that I would use it to find the minimum/minima point/points and choose values of $a$ such that those minima are $>0$. But the derivative I found is a whole mess with both $x's$ and $a's$ and I didn't know how to handle it.

Answer

$\frac {-a+|a|} 2$ is same as $\max (-a,0)$. Just consider the case $a \geq 0$ and $a<0$ to verify this.

The second question does not make sense. Whatever $a$ is, there are values of $x$ for which $f(x)$ is not defined.

However if the question is to find $a$ such that $f(x)>0$ for ll $x \in D$ then the answer is $a \geq 1$.

[We want $x+a >1$ whenever $x >\max (-a,0)$. It is easy to see that this will not hold if $a<0$. So take $a \geq 0$. Then the condition is $x>1-a$ whenever $x >0$. In particular this gives $\frac 1 n > 1-a$ for all $n \in \mathbb N$. Letting $n \to \infty$ we get $0 \geq 1-a$ or $a \geq 1$. Conversely if $a \geq 1$ it is easy to see that $f(x) >0$ for all $x \in D$].