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- Jan 26, 2012

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Another approach would be to use differentials:If \(\displaystyle f(x)\) is constant then I can show that \(\displaystyle f'(x)=0\).

But if \(\displaystyle f'(x)=0\), then how to show that \(\displaystyle f(x)\) is constant ?

$$\Delta y=f'(x)\, \Delta x = 0 \cdot \Delta x \, ...$$

- Jan 29, 2012

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The mean value theorem: if f is continuous and differentiable on [a, b] then there exist c in [a, b] such that [tex]f'(c)= \frac{f(b)- f(a)}{b- a}[/tex]. If f'(x)= 0 for all x, then for any a and b, [itex]\frac{f(b)- f(a)}{b- a}= 0[/tex] from which f(a)= f(b).If \(\displaystyle f(x)\) is constant then I can show that \(\displaystyle f'(x)=0\).

But if \(\displaystyle f'(x)=0\), then how to show that \(\displaystyle f(x)\) is constant ?