Is a function increasing if the derivative is positive except at one point of an interval? Let $f: \mathbb{R} \to \mathbb{R}$ be differentiable on $(a,b)$. Suppose $f' > 0$ on $(a,b)$ except at a point $c \in (a,b)$ (that is, $f'(c) \leq 0$).

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*Is $f$ increasing on $(a,b)$?

*Must $f'(c)$ be zero, or can it be negative?

Clearly $f$ is increasing on $(a,c) \cup (c,b)$ but I'm not sure about how the value at $c$ compares with the values at other points.
And I think $f'(c)$ must be zero: If $f'(c) < 0$ then for small positive $h$ we have $\frac{f(c+h) - f(c)}{h}$ is also negative (by definition of the derivative as a limit of this ratio), so $f(c+h) - f(c) < 0$. Since $f'$ is positive on $(c,c+h)$, the Mean Value Theorem implies that $f(c+h) - f(c) = f'(d)h$ for some $d \in (c, c+h)$, and $f'(d)h$ is a product of two positive numbers, hence positive. So $f(c+h) - f(c) > 0$, a contradiction.

You can use the fact that the derivative has intermediate value property to rule out $f'(c)<0$.(this will contradict that $c$ is only point where derivative is not positive)
Now pick a $x<c$ then by MVT there is $\eta\in(x,c)$ such that $f(c)-f(x)=f'(\eta)(c-x)>0$ [this is actually independent of derivative being defined at $c$ or not]. Similarly for $c<y$.
