A problem regarding the root of a continuously differentiable function on $\mathbb R$ $\mathbf {The \ Problem \ is}:$ If $f :\mathbb R \to \mathbb R$ is a continuously differentiable function such that inf $f'(x)=\delta(\gt 0).$ Then show that there exists a point $p \in \mathbb R$ such that $f(p)=0.$
$\mathbf {My \ approach} :$ Actually, here $f$ is a strictly increasing function on $\mathbb R$ and firstly if we let that $f(x) \geq 0$ for all $x \in \mathbb R$ and inf $f (x)=0$, then obviously $f$ shall never meet the $x$- axis and then there exists a sequence $x_n$ diverging to $-\infty$ such that $f(x_n) →0$ .
Then for than sequence $x_n$, I think (pictorially) $f'(x_n)→0$ using the continuity of $f'$, but I can't approach it .
And, note that in this case,as $f$ is continuous on $(-\infty,0]$ and $f(x)→0$ as $x→-\infty$, then $f$ is uniformly continuous on $(-\infty,0].$
A small hint is warmly appreciated .
 A: Ok, this is a hint, but not answer.
By the mean value theorem for continuos functions, it is sufficient to show that $f$ attains one negative and one positive value. (This should be clear). 
By the mean value theorem for differentiable functions, given $a$ and $b$ there exists $c$ (depending on $a$ and $b$) such that
$$f(a)-f(b) = (a-b) f^\prime(c)$$.
Now you know that $f^\prime(c)\ge \delta >0$
Fix $a$ and assume $f(a) >0$. Then check what this inequlity implies for $f(b)$, by making $b$ very large (positive and negative). Similarly if $f(b) < 0$.
A: Let $F(x) = F(a) + \int_a^x f'(z)dz>\delta (x-a)+F(a)>F(a)$ for all $a$ and $x>a$, so the function is strictly increasing. If there is some $a$ for which $F(a)<0$, then since $F$ is strictly increasing there is an interval on which it changes signs, and by the intermediate value theorem, there is a root somewhere in that interval.  But you seem to be thinking of something like $e^x$, for which the function and derivative go to zero as $x \rightarrow -\infty$, but $e^x$ has no root in $\mathbb{R}$.
A: If $f$ has  a local extremum at some point then $f'$ is $0$ at that point, which is a contradiction. Hence $f$ is strictly increasing on the whole line. If $f$ does not vanish at any point then it cannot change signs. Suppose $ f(x) >0$ for all $x$. Then $f(0)-f(-n) \geq \delta n$ by MVT so $f(0) \geq n\delta$ for every $n$! This contradiction finishes the proof in this case.  I will leave the case $f(x) <0$ for all $x$ to you. 
