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I wish to prove the following:

If $f: \mathbb{R} \to \mathbb{R}$ is a differentiable function, $x_0$ is a point in $\mathbb{R}$ such that $f'(x_0) = 0$, and $f''(x_0) > 0$ (so in particular $f''$ exists at $x_0$), then there is a $d > 0$ so that $f(x_0) < f(x)$ for all $x \in (x_0 - d, x_0 + d)$ that aren't equal to $x_0$.

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Hint: Use the definition of the derivative. If the limit as $h$ approaches $0$ of $\dfrac{f(x_0+h)-f(x_0)}{h}$ is positive, then $f(x_0+h)-f(x_0)$ is positive for all (non-zero) $h$ that are close enough to $0$. The notation of your question is a bit different, but this is exactly what you need.