My friend thinks the this is equal to $ f"(x_0) $, but I nothing see how to prove the this is true.

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Thanks for the help!

The easiest method to prove your friend wrong is to use this come a function that we really understand, choose $f(x) = x$. We"ll execute it at $x_0 = 1$. Then you room calculating

$$ fracf(1 + ah) - f(1 + bh)h = frac1 + ah - 1 - bhh = (a - b).$$

This is no $1$ and also depends critically top top $a$ and $b$ (exactly choose derivatives perform not).

More generally, we have the right to write (but where I omit all limit signs)$$ eginalignfracf(x_0 + ah) - f(x_0 + bh)h &= fracf(x_0 + ah) - f(x_0)h - fracf(x_0 + bh) - f(x_0)h \& o af"(x_0) - bf"(x_0) = (a-b)f"(x_0).endalign$$

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