Unique root to a function Let $f:[a,\infty)\rightarrow \mathbb{R}, \ \ f\in C^2[a,\infty)$ such that $$ \\ f(a)>0  , \ \ f'(a)<0, \ \ f''(x)\leq 0 \ \ \forall  x\in [a,\infty)$$ Prove that

$$ \exists !~t\in (a,\infty):f(t)=0$$

 A: From the mean value theorem, $f'(x)-f'(a)=(x-a)f''(\xi)$ for some $\xi\in(a,x)$, hence $f'(x)\le f'(a)<0$ for all $x\in(a,\infty)$ because $x-a\ge0$ and $f''(\xi)\le 0$. As a consequence $\frac{f'(x)}{f'(a)}\ge1$ for all $x\ge a$.
Let $x=a-\frac{f(a)}{f'(a)}$. Then $x>a$ and $f(x)-f(a)=(x-a)f'(\xi)=-f(a)\frac{f'(\xi)}{f'(a)}\le -f(a)$ because $f(a)>0$ and $\frac{f'(\xi)}{f'(a)}\ge1$.
In other words, $f(x)\le 0<f(a)$ and 
by the intermediate value theorem, there exists $t\in[a,x]$ with $f(t)=0$.
If there were two solutions $f(t_1)=f(t_2)=0$ with $t_1<t_2$, then by Rolle $f'(x)=0$ for some $x\in(t_1,t_2)$, contradicting $f'(x)<0$. Hence the solution is unique.
A: I will only give the idea which you can formalise yourself. It isn't too difficult, and should be a useful exercise.
The value of $f(a)$ is greater than zero, it's differential lower than zero. This means that, for $\epsilon$ sufficiently small, values $f(x')$ for $x'\in (a, \epsilon)$ will be lower than $f(a)$. 
If you take such an $x'$, you are in the same situation again, since all second derivatives negative or zero means that $f'(x')\le f'(a)$ for all $x'\in(a,\infty)$. So you know that, about $f(x')$, you can do the same trick as before.
The only difficulty then is to show that you can chose an epsilon small enough s.t. the trick works for any $x'\in (a,\infty)$ (i.e. chose an epsilon uniformly).
You can then iterate this, and since $\epsilon\gt 0$, you will reach $f(x')\le 0$ after finite iterations. The best part: you can continue doing this iteration to infininty, and your function will only decrease in value. 
This means you will have captures that one unique x', since it would otherwise be a contradiction to the intermediate value theorem.
Edit: Corrected terminology. I used mean-value instead of intermediate.
