# Sign of the first derivative

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 $$x_1 then $$f(x_1). I feel the statement contradicts one another. Any explanations will be appreciated.

• 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 $x_0$? – twosigma Jan 18 at 0:00
• @twosigma The second starement is also correct and it follows by definition of $f'(x_0)$. – Kavi Rama Murthy Jan 18 at 6:05
• @KaviRamaMurthy You are correct. – twosigma Jan 18 at 6:51

In the second statement you cannot take $$x_1$$ and $$x_2$$ on the same side of $$x_0$$. Hence there is no contradiction.
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).