There is not such function in $\mathbb{R}$ such that $f>0$, $f'>0$ and $f''<0$.

Question

Prove that there isn't a function $f \colon \mathbb{R} \to \mathbb{R}$ such that $f>0$, $f'>0$ and $f''<0$.

Any suggestion will be appreciated.

So the domain has to be $x \in \mathbb{R}$? Can it be a subset of $\mathbb{R}$? — Sheheryar Zaidi, Sep 14 '14 at 09:01
@Charlie: Being picky here, but $x=0$ gives 0, which means it is not strictly positive. Solution is simple, see my answer. — Sheheryar Zaidi, Sep 14 '14 at 09:08

score 1 · Answer 1 · answered Sep 14 '14 at 09:54

1

Since $f$ is strictly concave and non-constant, either $\lim_{x \to -\infty} f(x)=-\infty$ or $\lim_{x \to +\infty} f(x)=-\infty$. In both cases, $f$ can't be always positive.

answered Sep 14 '14 at 09:54

Siminore

35,136

Also, because f strictly increasing both limits must exists. – Jun 04 '16 at 10:54

score -1 · Answer 2 · answered Sep 14 '14 at 09:05

-1

$f(x)= \sqrt{x}+1$ is perhaps a counter example. It is always positive, has a positive derivative and concave down.

answered Sep 14 '14 at 09:05

Sheheryar Zaidi

2,714

1

The domain of f is real numbers. – user175913 Sep 14 '14 at 09:43

There is not such function in $\mathbb{R}$ such that $f>0$, $f'>0$ and $f''<0$.

2 Answers2