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Suppose $f,g\in C^1$ be continuously differentiable with the property that $f(x)\ge g(x)$ for all $x\in\mathbb R$ and $f(x_0)=g(x_0)$ for some $x_0\in\mathbb R$. Is it necessarily true that $f'(x_0)=g'(x_0)$?

Thanks

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1 Answer 1

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Consider $h(x) = f(x) - g(x), h(x)$ has a local minimum at $x_0.$

$f'(x_0) - g'(x_0) = 0$

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  • $\begingroup$ Thank you! Just wanted to check with another pair of eyes. $\endgroup$
    – Dosetsu
    Commented Sep 23, 2016 at 23:30

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