Minimum of $f(x)=\sqrt{9x^{2}+1}-3x-2 $ Let 
$$f(x)=\sqrt{9x^{2}+1}-3x-2 $$


*

*Show that $f$ is bounded from below by $-2$
$$\forall x\in \mathbb{R}\quad f(x)>-2 $$

*Is $-2$ the minimum value of $f$ ?


Indeed,
let $x\in \mathbb{R}$
\begin{align}
f(x)>-2 &\iff \sqrt{9x^{2}+1}-3x-2>-2\\
&\iff \sqrt{9x^{2}+1}>3x \\
&\implies 9x^{2}+1>9x^2 \\
&\implies 1>0
\end{align}
then $$\forall x\in \mathbb{R}\quad f(x)>-2 $$ which means $f$ is bounded from below by $-2$


*

*Is $-2$ the minimum value of $f$ ?


I can't show that $-2$ is not a value of f(x) to say that $-2$ isn't minimum value of $f$
Beware: No differentiability 
 A: Your $f(x)>-2$ argument is problematic because, as it is writtten, you can't go from $1>0$ in the bottom back to $f(x)>-2$. An easy fix is by replacing the last two $\Rightarrow$ by $\Leftarrow$'s. 
Alternatively, you can also rephrase the argument as follows: because $9x^2+1>9x^2$, we have
$$
f(x)>\sqrt{9x^2}-3x-2=3|x|-3x-2=3(|x|-x)-2\geq -2.
$$
And because the leftmost inequality is strict, $2$ is not the minimum value of $f$.
A: Note that 
$$
\sqrt{9x^2+1}>3x 
$$
if and only if one of the two conditions :
$$
x<0 \quad \mbox{or} \quad \begin{cases}
x\ge 0\\
9x^2+1>9x^2
\end{cases}
$$
is true. And, since $9x^2+1>9x^2$ is true $\forall x \in \mathbb{R}$, we can conclude that the function is $f(x)> -2 \quad \forall x \in \mathbb{R}$.
The same calculations shows that the equation $f(x)=-2$ has no solutions, so $-2$ is not a minimum for the function. 
A: We have :
$$f(x)=\sqrt{9x^{2}+1}-3x-2 $$
So
$$ f'(x) = \frac{9x}{\sqrt{9x^2+1}} - 3$$
It is obvious that $f'(x)\lt0$ :
$f'(x) \lt  0  \iff \frac{9x}{\sqrt{9x^2+1}} \lt  3 \iff \frac{81x^2}{9x^2+1} \lt 9   \iff 0\lt9$ . 
Now we know that $f(x)$ is strictly decreasing . Now we calculate limit using L'Hôpital :
$$ \lim_{x\to \infty}\sqrt{9x^{2}+1}-3x-2 = -2$$
Also we know $f(x) \neq -2$ as you calculate in your question Therefore we can conclude that $f(x)\gt -2$
