This function vanishes at some point Let $f:\mathbb{R}  \to \mathbb{R}$ be a continuously differentiable over $\mathbb{R}$ such that $ infimum \{ f' \left( x \right) \} >0$. Then prove that for some $a \in \mathbb{R}$ we have  $f\left(a\right)=0$
I am planning to consider two cases.
If function assume negative value at some point then because slope is positive and greater then $\delta>0$ it will keep on increasing forever.It cannot asymptotically converge to some  point.. hence it will take 0 at some point..
Similarly for other case by trending x to $-\infty$.
But I can't explain it mathematically.
Help me out
 A: Continuity of $f'$ is not needed.
Let $s=\inf f'(x)$. As in your idea, assume $f(0)<0$. Then for $a>0$,we have from the Mean Value Theorem that $f(a)-f(0)=af'(x)$ for some $x$ between $0$ and $a$. Therefore $f(a)\ge f(0)+as$. So if we pick $a>-\frac{f(0)}s$, we find $a$ with $f(a)>0$. Then by the Intermediate Value Theorem, $f(x)=0$ for some $x$ between $0$ and $a$. 
The same argument with negative $a$ works for the case $f(0)>0$.
A: Using the intermediate value theorem, you only need to show that there are some $a,b \in \mathbb{R}$, such that $f(a) < 0 < f(b)$.
I guess, you could use the mean value theorem. Let $a,b \in \mathbb{R}$, $a < b$. Then, there is some value $c \in [a,b]$ such that
$$f'(c) = \frac{f(b)-f(a)}{b-a} \Rightarrow \frac{f(b)-f(a)}{b-a} > \delta \Leftrightarrow f(b) > \delta(b-a)+f(a) \ \ \ (*)$$
On the right hand side you can fix $a$ and increase $b$ until $\delta \Leftrightarrow f(b) > \delta(b-a)+f(a)>0$
Similarly:
$$(*) \Leftrightarrow -f(a) > \delta(b-a)-f(b) \Leftrightarrow  f(a) < -\delta(b-a)+f(b),$$
where the right hand side can be made negative by fixing $b$ and choosing $a$ small enough to obtain $$f(a) < -\delta(b-a)+f(b) < 0.$$
