$\delta = \inf f'(x) > 0 $. prove that $f(a)= 0$ for some $a \in \mathbb{R} $. Let $ f: \mathbb{R} \rightarrow \mathbb{R}$ be a continuously differentiable function . Suppose $\delta = \inf f'(x) > 0 $. prove that $f(a)= 0$ for some $a \in \mathbb{R} $.
I have no idea how to proceed .I would like to share one observations
For any two points $a,b \in \mathbb{R}$ such that $f(a)=f(b)=0$ then we would have $f'(c)= 0$ for some $c \in (a,b)$ . So $0=f'(c) \geq \inf f'(x)= \delta > 0 $. Contradiction . Hence only one such point $a$ .
 A: If $f(0) = 0$ there is nothing to show. Otherwise let $a \in \mathbb R$ be nonzero. According to the mean-value theorem there exists $c$ in between $a$ and $0$ satisfying $$\frac{f(a) - f(0)}{a - 0} = f'(c) > \delta.$$
Assume first that $f(0) < 0$. Then for each $a > 0$ the above observation implies $f(a) > f(0) + a \delta$, and by taking $a$ sufficiently large you get $f(a) > 0$. Now apply the intermediate value theorem.
If $f(0) < 0$ take negative values of $a$ instead.
A: If
$d = \inf f'(x) > 0
$,
then
$f'(x) \ge d$
for all $x$.
Suppose $f(a) > 0$
for some $a$.
Then
$f(a)-f(a-x)
=\int_{a-x}^a f(t) dt
$
so
$\begin{array}\\
f(a-x)
&=f(a)-\int_{a-x}^a f'(t) dt\\
&\lt f(a)-\int_{a-x}^a d dt\\
&= f(a)-d(a-(a-x))\\
&= f(a)-dx\\
&=0
\text{ for } x = \frac{f(a)}{d}
\end{array}
$
Since $f$ is continuous,
$f(z) = 0$
for some
$a-\frac{f(a)}{d}
< z < a
$.
Similarly,
if $f(a) < 0$
for some $a$,
Then
$f(a+x)-f(a)
=\int_{a}^{a+x} f(t) dt
$
so
$\begin{array}\\
f(a+x)
&=f(a)+\int_{a}^{a+x} f'(t) dt\\
&\gt f(a)+\int_{a}^{a+x} d dt\\
&= f(a)+dx\\
&=0
\text{ for } x = \frac{-f(a)}{d}
\end{array}
$
Since $f$ is continuous,
$f(z) = 0$
for some
$a 
< z
<a-\frac{f(a)}{d}
$.
