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I'm trying to prove the following claim:

$f:\left[p,q\right]\:\to\mathbb R,\ f$ is differentiable at $p$ and $p$ is a minimum of $f$. prove or disprove: $f'(p) \ge 0$.

When I draw a graph, it seemed obvious that the claim is true. the minimum is at the left edge of the function, so the function must go up afterwards. but as I know, the derivative at minimum / maximum equals to $0$. So why am I being asked about $f'(p) \ge 0$ ?

Tryed to prove using rolle theorem, but I don't know if the function is differentiable at $q$ (and maybe it isn't).

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    Because you could have $f:[0,1]\to\Bbb R$ with $f(x)=x$. – 2017-01-23
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    By the way, Rolle's theorem does not require differentiability at the extremal points (just continuity), but in return it tells nothing about them. – 2017-01-23
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    is $f:\:\left[0,1\right]\:→\:R\:$ with $f(x) = x$ a counter-example? if so, i can't see why. – 2017-01-23
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    There is a technical caveat here: if $f$ is only defined on $[p,q]$, then $f$ can't be differentiable in the usual sense at $p$, because the definition of differentiable requires $f$ to be defined in a neighbourhood of $p$. But perhaps what is meant is that it is differentiable from the right at $p$. – 2017-01-23

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$$f'(p) = \lim_{x \to p \\ x > p} \frac{\overbrace{f(x) - f(p)}^{\ge 0}}{\underbrace{x-p}_{\ge 0}} \ge 0$$

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    what if we have $x\to p^-$? – 2017-01-23
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    @ΘΣΦGenSan $f$ is only defined on $[p,q]$. I presume that by "differentiable" at $p$, OP means "differentiable to the right" at $p$. – 2017-01-23
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    Opps, I forgot it. Thanks – 2017-01-23
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    I only miss an argument that the limit of a non-negative function is non-negative. – 2017-01-23
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    This might work: http://math.stackexchange.com/q/977068/86776 – 2017-01-23