# 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, $$f'(c) \leq 0$$).

• 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.

Let $$f\colon (a,b)\to \Bbb R$$ be continuous, and $$f'(x)>0$$ for $$x\in(a,b)\setminus\{c\}$$. We do not even need to assume that $$f'(c)$$ exists. Then $$f$$ is strictly increasing: Suppose $$a. Then $$f(x_1) follows from the Mean Value Theorem if $$x_2\le c$$ or if $$x_1\ge c$$. If $$x_1, just go in two steps via $$c$$.

Now suppose additionally that $$f'(c)=$$ exists. Then directly from the increasing property we get $$f'(c)\ge0$$.

• Just to add a bit more details: "in two steps via $c$": Since we are assuming continuity on $[x_1, c]$ and $[c, x_2]$, and differentiability on $(x_1, c)$ and $(c, x_2)$, applying the MVT to these two intervals gives $f(c)-f(x_1)>0$ and $f(x_2)-f(c) > 0$, so that $f(x_2) - f(x_1)=(f(x_2)-f(c))+(f(c)-f(x_1))>0$. "directly from the increasing property we get $f'(c)\geq 0$": Since $f'(c)$ exists, in particular $\lim_{h \to 0^+} \frac{f(c+h)-f(c)}{h}$ exists and equals $f'(c)$; the fraction $\frac{f(c+h)-f(c)}{h}$ has positive numerator and denominator, so the limit is $\geq 0$. Jul 28, 2020 at 18:01

According to Darboux's theorem, all functions that result from the differentiation of some other function on some interval have the intermediate value property (even though they need not be continuous).

Therefore $$f^\prime(c)=0$$.

From previous result, it follows that $$f$$ is indeed increasing on $$[a,b]$$ by applying the Mean Value Theorem.

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 $$x then by MVT there is $$\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.

• "this is actually independent of... or not". But, would you at least need to assume continuity at $c$? Jul 28, 2020 at 19:51
• @twosigma I think finitely many discontinuities are fine. Jul 28, 2020 at 20:31

If $$a and $$f$$ is continuous on $$(a,b)$$ and $$f'(x)>0$$ for $$x\in (a,c)\cup (c,b)$$ then $$f$$ is strictly increasing on $$(a,b)$$ regardless of whether or not $$f'(c)$$ even exists.

1. If $$(a or $$c and $$f(x)\ge f(y)$$ then since $$f$$ is differentiable on $$[x,y]$$ there exists $$z\in (x,y)$$ with $$f'(z)=\frac {f(y)-f(x)}{y-x}\le 0,$$ a contradiction.

2. If $$a and $$f(x)\ge f(c)$$ then since $$f$$ is continuous on $$[x,c]$$ there exists $$y\in (x,c)$$ with $$f(y)=\frac {1}{2}(\,f(x)+f(c)\,)\le f(x),$$ and since $$f$$ is differentiable on $$[x,y]$$ there exists $$z\in (x,y)$$ with $$f'(z)=\frac {f(y)-f(x)}{y-x}\le 0,$$ a contradiction.

3. The case $$c and $$f(c)\ge f(x)$$ is done similarly to 2. above.

Example: With $$a=0,c=1,b=2\,:$$

For $$x\in (0,1]$$ let $$f(x)=-\sqrt {1-x^2}\,.$$ For $$x\in [1,2)$$ let $$f(x)=\sqrt {1-(2-x)^2}\,.$$

• When you have written $f'(z) = \frac{f(y)-f(x)}{y-x} \leq 0$ in (1) and (2), did you mean $> 0$? Jul 28, 2020 at 19:45
• NO. In both cases $y>x$ and $f(y)\le f(x).$ Jul 28, 2020 at 23:32