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$f:[0,+\infty)\rightarrow \mathbb{R}$ is continuous function. s.t $f(0)=1$, $f(1)=0$ then existence of minimal $x>0$ s.t $f(x)=0$ is guaranteed?

If it has counterexample, for $C_1$ function $f$, is guaranteed the existence of minimal $x>0$ which satisfy above condition?

Please give me a proof or a counterexample for these.

Aryaman Maithani
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    Take $x=1$. Nothing more to be done. – Kavi Rama Murthy Feb 14 '21 at 05:59
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    $x=1$ is an example of a value of $x > 0$ for which $f(x) = 0$. There may be smaller positive values of $x$ for which $f(x) = 0$. It may be worthwhile to investigate whether if $a = \inf {x > 0: f(x) = 0}$, then $a \leq 1$ and $f(a)=0$. – leslie townes Feb 14 '21 at 06:02

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Since $f$ is continuous, the set $A = f^{-1}[\{0\}]$ is closed. It is bounded below since $A \subset [0, \infty)$ and thus, $c := \inf A$ exists and is an element of $A$.
Since $f(0) \neq 0$, $c \neq 0$ and thus, $c > 0$ is the minimal element you were looking for.

Aryaman Maithani
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