Given $f(x) \ge 0$ for every $x \in (a,b)$ and $c \in (a,b)$. If $f'(c)$ exist and $f(c) = 0$, proof $$f'(c) = 0.$$
My Attempt:
By using Mean Value Theorem: $$f'(c) = \frac{|f(b)-f(a)|}{|b-a|} \ge 0.$$ I want to proof that $f(b) - f(a)=0$. So that we can conclude that $f'(c) = 0$. But I dont get the way. Any hints?
Thanks in advance.
$\textbf{#1}$: The Mean Value Theorem is not really helping here, instead considering that $f(x)\geq 0\forall x\in (a,b)$ and $c\in (a.b)$. Given that $f(c)=0$, $c$ must be a local minima of the function in $(a,b)$. Therefore, the slope of the tangent of $f(x)$ at $x=c$, that is $f'(c)$, must be equal to $0$.
$\textbf{#2}$:Visualizing a graph of any such function also helps, if $f(x)\geq 0$ in $(a,b)$ and $f(c)=0$ where $c\in (a,b)$, then the $x$-axis must be tangent to the function at $c$, making $f'(c)=0$.