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The second-derivative test states that if $x$ is a real number such that $f'(x)=0$, then:

  1. If $f''(x)>0$, then $f$ has a local minimum at $x$.
  2. If $f''(x)<0$, then $f$ has a local maximum at $x$.
  3. If $f''(x)=0$, then the text is inconclusive.

I’m wondering whether non-standard analysis, involving the hyperreal numbers, allows for a more powerful version of the second-derivative test, one that has fewer or no inconclusive cases. In terms of nonstandard analysis, the second derivative of $f$ is defined as the standard part of $\frac{f’(x+\epsilon)-f’(x)}{\epsilon}$ for any infinitesimal $\epsilon$. But what if we omit the bit about the standard part? Fixing some infinitesimal $\epsilon$, let $g(x)=\frac{f’(x+\epsilon)-f’(x)}{\epsilon}$.

Then can we define a more powerful second derivative test as follows?

  1. If $g(x)>0$, then $f$ has a local minimum at $x$.
  2. If $g(x)<0$, then $f$ has a local maximum at $x$.
  3. If $g(x)=0$, then the text is inconclusive.

Is this valid, and would this test apply to a greater class of functions than the regular second derivative test?

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Let's take 1:

If $\frac{f’(x+\epsilon)-f’(x)}{\epsilon} > 0$ then $f$ has a local minimum at $x$.
False in general.1

This one works:

If $\frac{f’(x+\epsilon)-f’(x)}{\epsilon} > 0$ for all positive infinitesmial $\epsilon$, then $f$ has a local minimum at $x$.

But (by the transfer principle) that is the same as this one, using only ordinary real numbers $\epsilon$:

If $\frac{f’(x+\epsilon)-f’(x)}{\epsilon} > 0$ for all sufficiently small positive real $\epsilon$, then $f$ has a local minimum at $x$.

1 Example: $f(x) = \int_0^x t \sin \frac{1}{t}\;dt$, so $\frac{f’(0+\epsilon)-f’(0)}{\epsilon} = \sin\frac{1}{\epsilon}$, which is positive for some positive infinitesimal $\epsilon$ and negative for some other positive infinitesimal $\epsilon$.

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