I am confused by the statement that a read, that says $f'(x_0)>0$ does not imply that $f$ is increasing in an open interval around $x_0$. But the book also mentions that if $f(x)$ is differentiable at $x_0$ and $f'(x_0)>0$ then there is $h>0$ such that for all $x_1,x_2\in(x_0-h,x_0+h)$ if $x_1<x_0<x_2$ then $f(x_1)<f(x_0)<f(x_2)$. I feel the statement contradicts one another. Any explanations will be appreciated.
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2$\begingroup$ The first statement is right and the second statement is false. See math.stackexchange.com/questions/2768994/…. For the second statement, did the book not include the additional condition that $f'(x_0) > 0$ in an open interval around $x_0$? $\endgroup$ – twosigma Jan 18 at 0:00
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2$\begingroup$ @twosigma The second starement is also correct and it follows by definition of $f'(x_0)$. $\endgroup$ – Kavi Rama Murthy Jan 18 at 6:05
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$\begingroup$ @KaviRamaMurthy You are correct. $\endgroup$ – twosigma Jan 18 at 6:51
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In the second statement you cannot take $x_1$ and $x_2$ on the same side of $x_0$. Hence there is no contradiction.
answered Jan 18 at 0:02
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The first statement is correct: see twosigma's comment.
Adding to Kavi's answer, the second statement says that $f$ is strictly increasing at the point $x_0$, but it does not say that $f$ is increasing in an open interval around $x_0$ (which it isn't necessarily).
answered Jan 18 at 0:12
