Monotonic functions – "Math for Non-Geeks"

Monotonicity criterion

The monotonicity criterion is quite intuitive: if the derivative of a function (i.e. the slope) is positive, it goes up, if the derivative is negative, it goes down. Mathematically, if the derivative $f'$ of a differentiable function $f$ is non-negative or (non-positive) on an interval $(a,b)$ , then $f$ is monotonically increasing (or decreasing) on $(a,b)$ . If $f'$ is even strictly positive (or negative) $(a,b)$ , then $f$ is strictly monotonically increasing (or decreasing).

In the first case, even inversion of the statement is true: If a differentiable function is monotonically increasing on $[a,b]$ , then $f'(x)\geq 0$ and if the function is monotonically decreasing on $[a,b]$ , then then $f'(x)\geq 0$ . However, the inversion does not hold true in the strict case, monotone functions do not always have $f'(x)>0$ or $f'(x)<0$ . For instance, $f(x)=x^{3}$ is strictly monotone, but $f'(0)=3\cdot 0^{2}=0$ .

Theorem (Monotonicity criterion for differentiable functions)

Let $f:[a,b]\to \mathbb {R}$ be continuous and differentiable on $(a,b)$ . Then, there is

$f'\geq 0$ on $(a,b)$ $\iff$ $f$ monotonically increasing on $[a,b]$
$f'\leq 0$ on $(a,b)$ $\iff$ $f$ monotonically decreasing on $[a,b]$
$f'>0$ on $(a,b)$ $\Longrightarrow$ $f$ strictly monotonically increasing on $[a,b]$
$f'<0$ on $(a,b)$ $\Longrightarrow$ $f$ strictly monotonically decreasing on $[a,b]$

Proof

The four directions " $\Longrightarrow$ " follow from the mean value theorem. The two directions " $\Longleftarrow$ " follow by differentiability of the function:

Proof (monotonicity criterion for differentiable functions)

We first show the four directions " $\Longrightarrow$ " and then the two " $\Longleftarrow$ ".

1. $\Longrightarrow$ : From $f'\geq 0$ on $(a,b)$ we get that $f$ in monotonically increasing on $[a,b]$ .

Let $f'\geq 0$ for all $x\in (a,b)$ and let $x_{1},x_{2}\in [a,b]$ with $x_{1}\leq x_{2}$ . We need to show $f(x_{1})\leq f(x_{2})$ . By assumption, $f$ is continuous on $[x_{1},x_{2}]\subseteq [a,b]$ and differentiable on $(x_{1},x_{2})\subseteq (a,b)$ . By the mean-value theorem, there is a $\xi \in (x_{1},x_{2})$ with

${\frac {f(x_{2})-f(x_{1})}{x_{2}-x_{1}}}=f'(\xi )$

By assumption, $f'(\xi )\geq 0$ , and hence ${\tfrac {f(x_{2})-f(x_{1})}{x_{2}-x_{1}}}\geq 0$ . Since $x_{2}>x_{1}\iff x_{2}-x_{1}>0$ we have in the enumerator $f(x_{2})-f(x_{1})\geq 0$ . This is equivalent to $f(x_{2})\geq f(x_{1})$ , i.e. $f$ is monotonically increasing.

2. $\Longrightarrow$ : From $f'\leq 0$ on $(a,b)$ we get that $f$ in monotonically decreasing on $[a,b]$ .

Let $f'\leq 0$ for all $x\in (a,b)$ and let $x_{1},x_{2}\in [a,b]$ with $x_{1}\leq x_{2}$ . We need to show $f(x_{1})\geq f(x_{2})$ . By assumption, $f$ is continuous on $[x_{1},x_{2}]\subseteq [a,b]$ and differentiable on $(x_{1},x_{2})\subseteq (a,b)$ . By the mean-value theorem, there is a $\xi \in (x_{1},x_{2})$ with

${\frac {f(x_{2})-f(x_{1})}{x_{2}-x_{1}}}=f'(\xi )$

Now, $f'(\xi )\leq 0$ , and hence ${\tfrac {f(x_{2})-f(x_{1})}{x_{2}-x_{1}}}\geq 0$ . Since $x_{2}>x_{1}\iff x_{2}-x_{1}>0$ we have $f(x_{2})-f(x_{1})\leq 0$ . This is equivalent to $f(x_{2})\leq f(x_{1})$ , i.e. $f$ is monotonically decreasing.

3. $\Longrightarrow$ : $f'>0$ on $(a,b)$ implies that $f$ is strictly monotonically increasing on $[a,b]$

We prove this by contradiction: Let $f$ be not strictly monotonically increasing. That means, we have some $x_{1},x_{2}\in [a,b]$ with $x_{1}<x_{2}$ and $f(x_{1})\geq f(x_{2})$ . We need to find a $\xi \in (a,b)$ with $f'(\xi )\leq 0$ . Now, $f$ is continuous on $[x_{1},x_{2}]$ and differentiable on $(x_{1},x_{2})$ . So by the mean value theorem, we can find a $\xi \in (a,b)$ with

${\frac {f(x_{2})-f(x_{1})}{x_{2}-x_{1}}}=f'(\xi )$

Since $f(x_{1})\geq f(x_{2})$ , the enumerator of the quotient is non-positive, and because of $x_{1}<x_{2}$ the denominator is positive. Thus the whole fraction is non-positive, and therefore $f'(\xi )\leq 0$ .

4. $\Longrightarrow$ : $f'<0$ on $(a,b)$ implies that $f$ is strictly monotonically increasing on $[a,b]$

Another proof by contradiction: Let $f$ be not strictly monotonically decreasing. That means, we have some $x_{1},x_{2}\in [a,b]$ with $x_{1}<x_{2}$ and $f(x_{1})\leq f(x_{2})$ . We need to find a $\xi \in (a,b)$ with $f'(\xi )\geq 0$ . Now, $f$ is continuous on $[x_{1},x_{2}]$ and differentiable on $(x_{1},x_{2})$ . So by the mean value theorem, we can find a $\xi \in (a,b)$ with

${\frac {f(x_{2})-f(x_{1})}{x_{2}-x_{1}}}=f'(\xi )$

Since $f(x_{1})\leq f(x_{2})$ , the enumerator of the quotient is non-positive, and because of $x_{1}<x_{2}$ the denominator is positive. Thus the whole fraction is non-positive, and therefore $f'(\xi )\geq 0$ .

Now, the two directions " $\Longleftarrow$ " follow:

1. $\Longleftarrow$ : $f$ being monotonically increasing on $[a,b]$ implies $f'\geq 0$ on $(a,b)$

Let $x_{1},x_{2}\in [a,b]$ with $x_{1}<x_{2}$ . By monotonicity, $f(x_{1})\leq f(x_{2})$ . Further, let $x,{\tilde {x}}\in [x_{1},x_{2}]$ with $x\neq {\tilde {x}}$ . Then we have for the difference quotient

${\frac {f(x)-f({\tilde {x}})}{x-{\tilde {x}}}}\geq 0$

If $x>{\tilde {x}}$ , then $f(x)\geq f({\tilde {x}})$ . The enumerator and denominator of the difference quotient are thus non-negative, and so is the total quotient. Similarly in the case of $x<{\tilde {x}}$ and $f(x)\leq f({\tilde {x}})$ enumerator and denominator are non-positive. Thus the whole fraction is again non-negative. Now we form the differential quotient by taking the limit $x\to {\tilde {x}}$ . This limit exists because $f$ is differentiable on $(x_{1},x_{2})$ . Furthermore, the inequality remains valid because of the monotonicity rule for limit values. Thus we have

$f'({\tilde {x}})=\lim _{x\to {\tilde {x}}}{\frac {f(x)-f({\tilde {x}})}{x-{\tilde {x}}}}\geq 0$

Since $x_{1},x_{2}\in [a,b]$ and ${\tilde {x}}\in (x_{1},x_{2})$ have been arbitrary, we get $f'\geq 0$ on all of $(a,b)$ .

2. $\Longleftarrow$ : $f$ being monotonically decreasing on $[a,b]$ implies $f'\leq 0$ on $(a,b)$

Let again $x_{1},x_{2}\in [a,b]$ with $x_{1}<x_{2}$ . By monotonicity, $f(x_{1})\geq f(x_{2})$ . Further, let $x,{\tilde {x}}\in [x_{1},x_{2}]$ with $x\neq {\tilde {x}}$ . Then we have for the difference quotient

${\frac {f(x)-f({\tilde {x}})}{x-{\tilde {x}}}}\leq 0$

If $x>{\tilde {x}}$ , then $f(x)\leq f({\tilde {x}})$ and thus the total quotient is non-positive. An analogous statement holds in the case $x<{\tilde {x}}$ and $f(x)\geq f({\tilde {x}})$ . By forming the differential quotient we now obtain

$f'({\tilde {x}})=\lim _{x\to {\tilde {x}}}{\frac {f(x)-f({\tilde {x}})}{x-{\tilde {x}}}}\leq 0$

Since $x_{1},x_{2}$ and ${\tilde {x}}$ have been arbitrary, we get $f'\leq 0$ on all of $(a,b)$ .

Examples: monotonicity criterion

Quadratic and cubic functions

Example (Monotonicity of quadratic and cubic functions)

For the quadratic power function $f:\mathbb {R} \to \mathbb {R} ,\ f(x)=x^{2}$ there is

$f'(x)=2x\ {\begin{cases}>0&{\text{ if }}x>0,\\<0&{\text{ if }}x<0\end{cases}}$

So $f$ is strictly monotonically decreasing by the monotonicity criterion on $(-\infty ,0]$ and strictly monotonically increasing on $[0,\infty )$ .

For the cubic power function $g:\mathbb {R} \to \mathbb {R} ,\ g(x)=x^{3}$ there is

$g'(x)=3x^{2}\ {\begin{cases}\geq 0&{\text{ for all }}x\in \mathbb {R} ,\\>0&{\text{ for all }}x\in \mathbb {R} \setminus \{0\}\end{cases}}$

So by the monotonicity criterion, $g$ is monotonically increasing on $\mathbb {R}$ and strictly monotonically increasing on $(-\infty ,0]$ and $[0,\infty )$ . The cubic power function $g(x)=x^{3}$ is even strictly monotonically increasing on all of $\mathbb {R}$ .

The fact that $g$ with $g(x)=x^{3}$ is strictly' monotonically increasing, although only $f'\geq 0$ and not $f'>0$ , stems from its derivative being zero at only a single point (namely 0). In the end of this article, we will treat a criterion, which tells us when a function is strictly monotone, even if there is not everywhere $f'>0$ .

Question: Why is $g:\mathbb {R} \to \mathbb {R} :x\mapsto x^{3}$ strictly monotonically increasing on $\mathbb {R}$ ?

We must show: From $x,y\in \mathbb {R}$ with $x<y$ we get $g(x)<g(y)$ . For the cases $x<y\leq 0$ and $0\leq x<y$ we have already shown this with the monotonicity criterion. So we only have to look at the case $x<0<y$ . Here there is with the arrangement axioms (missing):

$g(x)=x^{3}=\underbrace {x} _{<0}\cdot \underbrace {x^{2}} _{>0}<0<\underbrace {y} _{>0}\cdot \underbrace {y^{2}} _{>0}=y^{3}=g(y)$

So $g$ is strictly monotonically increasing on all of $\mathbb {R}$ .

Warning

In the example $x\mapsto x^{3}$ we have seen that the statement " $f'>0$ implies strict monotonicity" does not hold true! This means that from the fact that $f$ increases strictly monotone, we can in general not conclude that $f'>0$ . In the example of the function $h:\mathbb {R} \to \mathbb {R} :x\mapsto -x^{3}$ one can also see that the statement " $f'<0$ implies strictly monotone falling" does not hold true in general.

Exponential and logarithm function

Example (Monotonicity of the exponential and logarithm function)

For the exponential function $f:\mathbb {R} \to \mathbb {R} ,\ f(x)=\exp(x)$ there is for all $x\in \mathbb {R}$ :

$f'(x)=\exp(x)>0$

Therefore, according to the monotonicity criterion, $\exp$ is strictly monotonically increasing on all of $\mathbb {R}$ . For the (natural) logarithm function $g:\mathbb {R} ^{+}\to \mathbb {R} ,\ g(x)=\ln(x)$ there is for all $x\in \mathbb {R} ^{+}$ :

$g'(x)={\frac {1}{x}}>0$

So $\ln$ is strictly monotonically increasing on $\mathbb {R} ^{+}$ (not including the 0).

Question: What is the monotonicity behaviour of the logarithm function extended to $\mathbb {R} ^{-}$ , i.e. $:\mathbb {R} \setminus \{0\}\to \mathbb {R} ,\ h(x)=\ln |x|$ ?

There is

$h(x)={\begin{cases}\ln(x)&{\text{ for }}x>0\\\ln(-x)&{\text{ for }}x<0\end{cases}}$

Above we have shown that $h'(x)>0$ for $x\in \mathbb {R} ^{+}$ . So $h$ is strictly monotonically increasing on $\mathbb {R} ^{+}$ , as well . For $x<0$ on the other hand there is $h'(x)={\frac {1}{-x}}\cdot (-1)={\frac {1}{x}}<0$ . So $h$ is strictly monotonically decreasing on $\mathbb {R} ^{-}$ .

Trigonometric functions

Example (Monotonicity of the sine function)

For the sine function $f:\mathbb {R} \to \mathbb {R} ,\ f(x)=\sin(x)$ there is

$f'(x)=\cos(x)\ {\begin{cases}>0&{\text{ for }}x\in (-{\tfrac {\pi }{2}}+2\pi k,{\tfrac {\pi }{2}}+2\pi k),\ k\in \mathbb {Z} ,\\<0&{\text{ for }}x\in ({\tfrac {\pi }{2}}+2\pi k,{\tfrac {3\pi }{2}}+2\pi k),\ k\in \mathbb {Z} \end{cases}}$

So for all $k\in \mathbb {Z}$ , the $\sin$ is strictly monotonically increasing on the intervals $[-{\tfrac {\pi }{2}}+2\pi k,{\tfrac {\pi }{2}}+2\pi k]$ and strictly monotonically decreasing on the intervals $[{\tfrac {\pi }{2}}+2\pi k,{\tfrac {3\pi }{2}}+2\pi k]$ .

Question: Where does the cosine function $g:\mathbb {R} \to \mathbb {R} ,\ g(x)=\cos(x)$ show monotone behaviour?

Here, $g'(x)=-\sin(x)\ {\begin{cases}>0&{\text{ for }}x\in (\pi +2\pi k,2\pi +2\pi k),\ k\in \mathbb {Z} ,\\<0&{\text{ for }}x\in (2\pi k,{\tfrac {3\pi }{2}}+\pi +2\pi k),\ k\in \mathbb {Z} \end{cases}}$ .

So for all $k\in \mathbb {Z}$ , the $\cos$ is strictly monotonically increasing on the intervals $[\pi +2\pi k,2\pi +2\pi k]$ and strictly monotonically decreasing on the intervals $[2\pi k,\pi +2\pi k]$ .

Example (Monotonicity of the tangent function)

For the tangent function $\tan :D=\mathbb {R} \setminus \{{\tfrac {\pi }{2}}+k\pi \mid k\in \mathbb {Z} \}\to \mathbb {R} ,\ \tan(x)={\tfrac {\sin(x)}{\cos(x)}}$ there is for all $x\in D$ :

$\tan '(x)=1+\underbrace {\tan ^{2}(x)} _{\geq 0}\geq 1>0$

Hence, for all $k\in \mathbb {Z}$ , the $\tan$ is strictly monotonically increasing on the intervals $(-{\tfrac {\pi }{2}}+k\pi ,{\tfrac {\pi }{2}}+k\pi )$ .

Question: Where does the cotangent function $\cot :{\tilde {D}}=\mathbb {R} \setminus \{k\pi \mid k\in \mathbb {Z} \}\to \mathbb {R} ,\ \cot(x)={\tfrac {\cos(x)}{\sin(x)}}$ show monotone behaviour?

For all $x\in {\tilde {D}}$ , there is

$\cot '(x)=-1-\underbrace {\tan ^{2}(x)} _{\geq 0}\leq -1<0$

So for all $k\in \mathbb {Z}$ , the $\cot$ is strictly monotonically decreasing on the intervals $(k\pi ,\pi +k\pi )$ .

Exercise

Monotonicity intervals and existence of a zero

Exercise (Monotonicity intervals and existence of a zero)

Where is the following polynomial function monotone?

$f:\mathbb {R} \to \mathbb {R} ,\ f(x)=x^{3}-x-1$

Prove that $f$ has exactly one zero.

Solution (Monotonicity intervals and existence of a zero)

Monotonicity intervals:

The function $f$ is differentiable on all of $\mathbb {R}$ , with

$f'(x)=3x^{2}-1=3\left(x^{2}-{\tfrac {1}{3}}\right)=3\left(x+{\tfrac {1}{\sqrt {3}}}\right)\left(x-{\tfrac {1}{\sqrt {3}}}\right)$

So

$f'(x)>0\iff {\begin{cases}x+{\tfrac {1}{\sqrt {3}}}>0{\text{ and }}x-{\tfrac {1}{\sqrt {3}}}>0\iff x>-{\tfrac {1}{\sqrt {3}}}{\text{ and }}x>{\tfrac {1}{\sqrt {3}}}\iff x>{\tfrac {1}{\sqrt {3}}}\\x+{\tfrac {1}{\sqrt {3}}}<0{\text{ and }}x-{\tfrac {1}{\sqrt {3}}}<0\iff x<-{\tfrac {1}{\sqrt {3}}}{\text{ and }}x<{\tfrac {1}{\sqrt {3}}}\iff x<-{\tfrac {1}{\sqrt {3}}}\end{cases}}$

According to the monotonicity criterion, $f$ is strictly monotonically increasing on $\left(-\infty ,-{\tfrac {1}{\sqrt {3}}}\right)$ and on $\left({\tfrac {1}{\sqrt {3}}},\infty \right)$ . Further,

$f'(x)<0\iff {\begin{cases}x+{\tfrac {1}{\sqrt {3}}}>0{\text{ and }}x-{\tfrac {1}{\sqrt {3}}}<0\iff x>-{\tfrac {1}{\sqrt {3}}}{\text{ and }}x<{\tfrac {1}{\sqrt {3}}}\iff -{\tfrac {1}{\sqrt {3}}}<x<{\tfrac {1}{\sqrt {3}}}\\x+{\tfrac {1}{\sqrt {3}}}<0{\text{ and }}x-{\tfrac {1}{\sqrt {3}}}>0\iff x<-{\tfrac {1}{\sqrt {3}}}{\text{ and }}x>{\tfrac {1}{\sqrt {3}}}\iff {\text{ not possible}}\end{cases}}$

According to the monotonicity criterion, $f$ is strictly monotonically decreasing on $\left(-{\tfrac {1}{\sqrt {3}}},{\tfrac {1}{\sqrt {3}}}\right)$ .

$f$ has exactly one zero:

For $f$ , we have the following table of values:

${\begin{array}{|c||c|c|c|c|c|c|}\hline x&-1&-{\tfrac {1}{\sqrt {3}}}&0&{\tfrac {1}{\sqrt {3}}}&1&2\\\hline f(x)&-1&-1+{\tfrac {2}{3{\sqrt {3}}}}&-1&-1-{\tfrac {2}{3{\sqrt {3}}}}&-1&5\\\hline \end{array}}$

Based on the monotonicity properties and the continuity of $f$ that we have previously investigated, we can read off that:

On $\left(-\infty ,-{\tfrac {1}{\sqrt {3}}}\right)$ is $f$ strictly monotonically increasing. Because of $f\left(-{\tfrac {1}{\sqrt {3}}}\right)=-1+{\tfrac {2}{3{\sqrt {3}}}}<0$ there is $f(x)<0$ for all $x\in \left(-\infty ,-{\tfrac {1}{\sqrt {3}}}\right)$ .
On $\left(-{\tfrac {1}{\sqrt {3}}},{\tfrac {1}{\sqrt {3}}}\right)$ is $f$ then strictly monotonically decreasing. So there is also $f(x)<0$ for all $x\in \left(-{\tfrac {1}{\sqrt {3}}},{\tfrac {1}{\sqrt {3}}}\right)$ .
Subsequently $f$ increases on $\left({\tfrac {1}{\sqrt {3}}},\infty \right)$ again strictly monotonically. Because of $f(1)=-1<0$ and $f(2)=5>0$ , there must be an $x_{0}\in (1,2)$ with $f(x_{0})=0$ by the mean value theorem. Because of the strict monotonicity of $f$ on $\left({\tfrac {1}{\sqrt {3}}},\infty \right)$ , there cannot be any further zeros.

Necessary and sufficient criterion for strict monotonicity

Exercise (Necessary and sufficient criterion for strict monotonicity)

Prove that: a continuous function $f:[a,b]\to \mathbb {R}$ which is differentiable to $(a,b)$ is strictly monotonically increasing exactly when there is

$f'(x)\geq 0$ for all $x\in (a,b)$
The zero set of $f'$ contains no open interval.

As an application: Show that the function $f:\mathbb {R} \to \mathbb {R} ,f(x)=x-\sin(x)$ is strictly monotonically increasing on all of $\mathbb {R}$ .

Proof (Necessary and sufficient criterion for strict monotonicity)

From the monotonicity criterion we already know that $f$ is monotonically increasing exactly when $f'(x)\geq 0$ . So we only have to show that $f$ is strictly monotonically increasing exactly when the second condition is additionally fulfilled.

$\Longrightarrow$ : $f$ strictly monotonically increasing $\Longrightarrow$ the set of zeros of $f'$ does not contain an open interval.

We perform a proof by contradiction. In other words, we show: If the set of zeros of $f'$ contains an open interval, $f$ is not strictly monotonically increasing. Assume there is $a_{1},b_{1}\in (a,b)$ with $f'(x)=0$ for all $x\in (a_{1},b_{1})$ . Then, by the mean value theorem there is a $\xi \in (a_{1},b_{1})$ with

$f(b_{1})-f(a_{1})=\underbrace {f'(\xi )} _{=0}(b_{1}-a_{1})=0$

So $f(a_{1})=f(b_{1})$ . If now $x\in (a_{1},b_{1})\iff a_{1}<x<b_{1}$ , then since $f$ is monotonically increasing, there is

$f(a_{1})\leq f(x)\leq f(b_{1})$

So there is $f(x)=f(b_{1})$ for all $x\in (a_{1},b_{1})$ . Hence, $f$ is not strictly monotonically increasing.

$\Longleftarrow$ : the set of zeros of $f'$ does not contain an open interval $\Longrightarrow$ $f$ strictly monotonically increasing

We perform a proof by contradiction. In other words, we show: if $f$ is monotonically, but not strictly monotonically increasing, then the zero set of $f'$ contains an open interval. Assume there is $a_{2},b_{2}\in (a,b)$ with $a_{2}<b_{2}$ with $f(a_{2})=f(b_{2})$ . Because of the monotonicity of $f$ there is

$f(a_{2})\leq f(x)\leq f(b_{2})$

So $f(x)=f(b_{2})$ for all $x\in (a_{1},b_{1})$ . That means $f$ is constant on $(a_{2},b_{2})$ . Hence there is for all $x\in (a_{2},b_{2})$ :

$f'(x)=0$

so the set of zeros of $f'$ does contain an open interval and we get a contradiction.

Exercise: $f:\mathbb {R} \to \mathbb {R} ,\ f(x)=x-\sin(x)$ is strictly monotonically increasing

$f$ is differentiable for all $x\in \mathbb {R}$ where

$f'(x)=1-\cos(x)\geq 0$

as $\cos(x)\leq 1$ for all $x\in \mathbb {R}$ . Hence $f$ is monotonically increasing. Further there is

${\begin{aligned}&f'(x)=1-\cos(x)=0\\\iff {}&\cos(x)=1\\\iff {}&x\in \{2\pi k\mid k\in \mathbb {Z} \}\end{aligned}}$

So the set of zeros of $f'$ contains only isolated points, and thus no open interval. Therefore $f$ is strictly monotonically increasing on $\mathbb {R}$ .