Is a function increasing if the derivative is positive except at one point of an interval?

Question

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.

Answer 1

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 $a<x_1<x_2<b$. Then $f(x_1)<f(x_2)$ follows from the Mean Value Theorem if $x_2\le c$ or if $x_1\ge c$. If $x_1<c<x_2$, just go in two steps via $c$.

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

Answer 2

Answer to question 2

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

Answer to question 1

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

Answer 3

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 $c<y$.

Answer 4

If $a<c<b$ 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 $(a<x<y<c$ or $c<x<y<b)$ 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 $a<x<c$ 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 $c<x<b$ 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 $f(x)=\sqrt {1-(2-x)^2}\,.$