Does there exist a real function with domain $\Bbb{R}$ such that $f'(x)>0$ and $f''(x)+(f'(x))^2<0$ for all $x$?

Question

Does there exist a real function $f(x)$ that satisfies the following properties?

1. its domain is $\mathbb{R}$

2. $f'(x) > 0$ for all $x$

3. $f''(x) + (f'(x))^2 < 0$ for all $x$

The log function $\ln(x)$ gives some idea about conditions 2 and 3. But for now, I did not find any example.

Besides, I want to find a non-linear differential function $f(x)$ defined on $\mathbb{R}$ and that is:

1. strictly increasing

2. (quasi)-concave

I think this is easier than the previous one. By looking at the graph, I guess this function exists, but I did not find one explicitly.

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EDIT 1

Thank you for all the comments, especially by @mihaild. I have found an example for these questions. An example for the 2nd question is $-e^{-x}$. And an example for the 1st question is borrowing the idea of the 2nd one, which is the following.

If $f(x)=-e^{g(x)}$, then $f'(x)=-g'(x)e^{g(x)}$, $f''(x)=-g''(x)e^{g(x)}-(g'(x))^2e^{g(x)}$, and $f''(x)+(f'(x))^2=-g''(x)e^{g(x)}$. So we just need $g(x)$ is defined on $\mathbb{R}$ and satisfies:

1. $g'(x) <0$ for all $x$,

2. $g''(x) >0$ for all $x$

Then simply we choose $g(x)=e^{-x}$. So an example is $-e^{e^{-x}}$.

This is a wrong computation. I will try to fix it. $(f’(x))^2$ should be $(g’(x))^2 e^{2g(x)}$.

EDIT 2

After checking my previous example and reading all comments again. I see there does not exist a real function for the 1st question. (Thanks a lot for a remark in @mihaild comment.)

Indeed, assume that there exists $f(x)$ that satisfies the 1st condition. Then $g(x)=e^{f(x)}$ is a positive, strictly increasing and strictly concave function. But by its concavity $$g(x) \leq g(0)+ g'(0)x.$$ Since $g'(0)>0$, we have $\displaystyle\lim_{x \to - \infty}g(x) =-\infty$, which contradicts its positivity.

Answer 1

Let $g=1/f'$ so $g'>1,g>0\,\forall x\in \Bbb R$ which is impossible so such an $f$ does not exist.

Answer 2

Adding to TheSimpliFire's solution (because I cannot comment), one can prove the assertion that a differentiable function from $\mathbb{R}$ to itself such that $g>0$ and $g'>1$ cannot exist by applying the mean value theorem to $g$, giving \begin{equation} |g(x)-g(y)|>|x-y| \text{ for all x,y}. \end{equation} By definition, the function is injective. Being continuous and injective it takes open sets to open sets. So $f(\mathbb{R})$ is open. Also, taking a sequence $(x_{n})$ converging to $x$ in $\mathbb{R}$ we observe that $f(x_{n})$ goes to $f(x)$ so $f(\mathbb{R})$ is closed. Being both open and closed in $\mathbb{R}$, $f(\mathbb{R}) = \mathbb{R}$.

The function is surjective, contradicting the fact that it can only take positive values.

Answer 3

After TheSimpleFire's elegant solution. Is my reasoning simple and correct?:

Let $g(0)=N>0$. Then $$\int_{-N-1}^0 g'(x)dx>\int_{-N-1}^0 1\;dx$$ $$g(0)-g(-N-1)>N+1$$ $$N-g(-N-1)>N+1$$ $$g(-N-1)<-1$$ contradicting the positivity of $g(x)$.