Show that $f(0)>0,\; f'(0)=0,\; f''(x)<0$ imply that $f(x)=0$ has exactly 1 positive root. 
Let $f$ be a function twice differentiable on $\mathbb{R}$. Suppose that $f(0)>0, \; f'(0)=0$ and $\; f''(x)<0$ for all $x>0$. Prove that $f(x)=0$ has exactly one positive root.

We have $f''(x)<0 \implies f'(x)$ is strictly decreasing on $(0,\infty)$. Then $f'(0)=0$, so $f'(x)<0$ on $(0,\infty)$. So $f$ has at most 1 positive root.
$f'(x)<0 \implies f(x)$ is strictly decreasing on $(0,\infty)$. 
How do I make use of the fact that $f(0)>0$ to show $f(x)=0$ has at least 1 positive root?
 A: The condition $f''<0$ implies that $f'$ is strictly decreasing. Since $f'(0)=0$, $f'(a)$ must be negative for some positive $a$ that is close to zero. But then, as $f'$ is strictly decreasing, $f'(x)<f'(a)$ for every $x\ge a$. By mean value theorem, when $x>a$, $f(x)-f(a)=f'(c)(x-a)$ for some $c\in(x,a)$. Hence $f(x)<f(a)+f'(a)(x-a)$ whenever $x>a$, meaning that $f(x)$ is eventually negative when $x\to+\infty$. Hence $f(b)<0$ for some $b>a$. Now $f(0)>0>f(b)$. By intermediate value theorem, $f(x)=0$ has a positive root. As $f$ is strictly decreasing, this root is unique.
A: Hint. Since $f$ is strictly concave in $(0,+\infty)$, we have that for $x_0,x>0$, 
$$f(x)\leq f'(x_0)(x-x_0)+ f(x_0).$$
that is the graph of $f$ stays under its tangent at $x_0$. 
Note that here $f'(x_0)<0$. Take the limit as $x\to +\infty$. What may we conclude?
A: You may show this also as follows:


*

*Uniqueness: Assume $x_1,x_2 > 0$ with $f(x_1) = f(x_2) = 0$ and $x_1 \neq x_2$
$$\Rightarrow \exists \xi > 0: f'(\xi) = 0 \mbox{ contradiction to } f' \mbox{ strictly decreasing and } f'(0) = 0.$$

*Existence: Consider $x \geq 1$ and use MVT for continuous functions:
$$f(x) = f(1) + \int_{1}^x f'(t)\;dt \stackrel{f'(t) \leq f'(1) <0}{\leq} f(1) + (x-1)\cdot \underbrace{f'(1)}_{<0} \color{blue}{<0} \mbox{ for } x \mbox{ large enough.}$$
So, $f$ changes the sign on $[0,\infty)$.

A: Since the second derivative $f''$ is negative in $(0,\infty) $, the first derivative $f' $ is strictly decreasing in $[0,\infty)$. And we have $f'(0)=0$ so that $f'$ is negative in $(0,\infty)$. It thus follows that $f'(x) $ either tends to a negative limit or diverges to $-\infty $ as $x\to\infty$.
The fact that derivative $f'$ is negative leads us to the conclusion that $f$ is strictly decreasing on $[0,\infty)$ and hence $f(x) $ either tends to a limit or diverges to $-\infty $ as $x\to\infty $. If $f(x) \to L$ then by mean value theorem we have $$f(x+1)-f(x)=f'(\xi)$$ for some $\xi\in(x, x+1)$. Now as $x\to\infty$ the LHS of above equation tends to $L-L=0$ and the RHS remains strictly away from $0$ (see last paragraph) which is a contradiction. Thus $f(x) \to-\infty$ as $x\to\infty $. Since $f(0)>0$, it follows by intermediate value theorem that $f(x) $ vanishes at least once in $(0,\infty) $ and since $f$ is strictly monotone it vanishes only once. Thus $f$ has exactly one positive real root. 
