Is there a solution to this problem? I have been thinking about this for some time, but haven't hit upon anything yet. Any help will be appreciated.
Is there a function mapping $\mathbf R$ into $\mathbf R$  satisfying the following: It is entirely differentiable on the real line, everywhere positive and its derivative decreases as $x$ (the argument) increases.
If it is impossible, why?
I have mostly just tried to think of combinations of functions I know, but without luck. Also, I tried to show its impossibility, but have come up with nothing so far.
 A: No such function exists, unless you allow non-strictly decreasing derivative and thereby accept constant functions.
If a differentiable function $f: \mathbb{R} \to \mathbb{R}$ has a decreasing derivative, then $f'(x)$, being monotonic, either approaches a constant or $\pm \infty$ as $x \to \pm \infty$. If $f$ is to be everywhere positive, the $x \to - \infty$ limit of $f'(x)$ must be non-positive, while the $x \to \infty$ limit of $f'(x)$ must be non-negative, so the derivative cannot be decreasing unless it is identically zero.
A: As it turns out, such a function does not exist!
To see this, we show that any differentiable $f: \mathbb{R} \rightarrow \mathbb{R}$ with $f'$ (strictly) decreasing on all of $\mathbb{R}$ cannot be entirely positive. So let $f: \mathbb{R} \rightarrow \mathbb{R}$ be differentiable and $f'$ be decreasing. Start by noting that since $f'$ is decreasing, we may take $f'(0) \neq 0$ (by translating $f$ if necessary). Now, if $f'(0) < 0$, then at $k = - 2\frac{f(0)}{f'(0)}$, the fact that $f'$ is decreasing shows that $$f(k) = f(0) + \int_0^k{f'(t)}dt < f(0) + kf'(0) = f(0) - 2f(0) \leq 0.$$ Similarly, if $f'(0) \geq 0$, we can apply a symmetric argument to find a value $k < 0$ with $f(k) < 0$. This completes our proof.
