Monotonic functions

Monotony criterion

The monotony 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 $f^{\prime }$ of a differentiable function $f$ is non-negative or (non-positive) on an interval $(a,b)$ , then $f$ is monotonously increasing (or decreasing) on $(a,b)$. If $f^{\prime }$ is even strictly positive (or negative) $(a,b)$, then $f$ is strictly monotonously increasing (or decreasing).

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

Theorem (Monotony criterion for differentiable functions)

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

1. $f^{\prime }\ge 0$ on $(a,b)$ $⟺$ $f$ monotonously increasing on $[a,b]$
2. $f^{\prime }\le 0$ on $(a,b)$ $⟺$ $f$ monotonously decreasing on $[a,b]$
3. $f^{\prime }>0$ on $(a,b)$ $⟹$ $f$ strictly monotonously increasing on $[a,b]$
4. $f^{\prime }<0$ on $(a,b)$ $⟹$ $f$ strictly monotonously decreasing on $[a,b]$

Proof

The four directions "$⟹$" follow from the mean value theorem. The two directions "$⟸$" follow by differentiability of the function:

Proof (monotony criterion for differentiable functions)

We first show the four directions "$⟹$" and then the two "$⟸$".

1. $⟹$: From $f^{\prime }\ge 0$ on $(a,b)$ we get that $f$ in monotonously increasing on $[a,b]$.

Let $f^{\prime }\ge 0$ for all $x\in (a,b)$ and let $x_1,x_2\in [a,b]$ with $x_1\le x_2$. We need to show $f(x_1)\le f(x_2)$. 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

$\frac{f(x_2)-f(x_1)}{x_2-x_1}=f^{\prime }(\xi )$

By assumption, $f^{\prime }(\xi )\ge 0$, and hence $\frac{f(x_2)-f(x_1)}{x_2-x_1}\ge 0$. Since $x_2>x_1⟺x_2-x_1>0$ we have in the enumerator $f(x_2)-f(x_1)\ge 0$. This is equivalent to $f(x_2)\ge f(x_1)$, i.e. $f$ is monotonously increasing.

2. $⟹$: From $f^{\prime }\le 0$ on $(a,b)$ we get that $f$ in monotonously decreasing on $[a,b]$.

Let $f^{\prime }\le 0$ for all $x\in (a,b)$ and let $x_1,x_2\in [a,b]$ with $x_1\le x_2$. We need to show $f(x_1)\ge f(x_2)$. 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

$\frac{f(x_2)-f(x_1)}{x_2-x_1}=f^{\prime }(\xi )$

Now, $f^{\prime }(\xi )\le 0$, and hence $\frac{f(x_2)-f(x_1)}{x_2-x_1}\ge 0$. Since $x_2>x_1⟺x_2-x_1>0$ we have $f(x_2)-f(x_1)\le 0$. This is equivalent to $f(x_2)\le f(x_1)$, i.e. $f$ is monotonously decreasing.

3. $⟹$: $f^{\prime }>0$ on $(a,b)$ implies that $f$ is strictly monotonously increasing on $[a,b]$

We prove this by contradiction: Let $f$ be not strictly monotonously increasing. That means, we have some $x_1,x_2\in [a,b]$ with $x_1<x_2$ and $f(x_1)\ge f(x_2)$. We need to find a $\xi \in (a,b)$ with $f^{\prime }(\xi )\le 0$ . 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

$\frac{f(x_2)-f(x_1)}{x_2-x_1}=f^{\prime }(\xi )$

Since $f(x_1)\ge f(x_2)$ , 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 $f^{\prime }(\xi )\le 0$.

4. $⟹$: $f^{\prime }<0$ on $(a,b)$ implies that $f$ is strictly monotonously increasing on $[a,b]$

Another proof by contradiction: Let $f$ be not strictly monotonously decreasing. That means, we have some $x_1,x_2\in [a,b]$ with $x_1<x_2$ and $f(x_1)\le f(x_2)$. We need to find a $\xi \in (a,b)$ with $f^{\prime }(\xi )\ge 0$ . 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

$\frac{f(x_2)-f(x_1)}{x_2-x_1}=f^{\prime }(\xi )$

Since $f(x_1)\le f(x_2)$ , 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 $f^{\prime }(\xi )\ge 0$.

Now, the two directions "$⟸$" follow:

1. $⟸$: $f$ being monotonously increasing on $[a,b]$ implies $f^{\prime }\ge 0$ on $(a,b)$

Let $x_1,x_2\in [a,b]$ with $x_1<x_2$. By monotony, $f(x_1)\le f(x_2)$. Further, let $x,\tilde{x}\in [x_1,x_2]$ with $x\ne \tilde{x}$. Then we have for the difference quotient

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

If $x>\tilde{x}$, then $f(x)\ge f(\tilde{x})$. 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 $f(x)\le f(\tilde{x})$ 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 monotony rule for limit values. Thus we have

$f^{\prime }(\tilde{x})=\underset{x\to \tilde{x}}{lim}\frac{f(x)-f(\tilde{x})}{x-\tilde{x}}\ge 0$

Since $x_1,x_2\in [a,b]$ and $\tilde{x}\in (x_1,x_2)$ have been arbitrary, we get $f^{\prime }\ge 0$ on all of $(a,b)$.

2. $⟸$: $f$ being monotonously decreasing on $[a,b]$ implies $f^{\prime }\le 0$ on $(a,b)$

Let again $x_1,x_2\in [a,b]$ with $x_1<x_2$. By monotony, $f(x_1)\ge f(x_2)$. Further, let $x,\tilde{x}\in [x_1,x_2]$ with $x\ne \tilde{x}$. Then we have for the difference quotient

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

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

$f^{\prime }(\tilde{x})=\underset{x\to \tilde{x}}{lim}\frac{f(x)-f(\tilde{x})}{x-\tilde{x}}\le 0$

Since $x_1,x_2$ and $\tilde{x}$ have been arbitrary, we get $f^{\prime }\le 0$ on all of $(a,b)$.

Examples: monotony criterion

Quadratic and cubic functions

Example (Monotony of quadratic and cubic functions)

Graphs of the functions $f$ and $g$

Graphs of the functions $f^{\prime }$ and $g^{\prime }$

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

$f^{\prime }(x)=2x\{\begin{array}{cc}>0& \text{ if }x>0,\\ <0& \text{ if }x<0\end{array}$

So $f$ is strictly monotonously decreasing by the monotony criterion on $(-\mathrm{\infty },0]$ and strictly monotonously increasing on $[0,\mathrm{\infty })$ .

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

$g^{\prime }(x)=3x^2\{\begin{array}{cc}\ge 0& \text{ for all }x\in ℝ,\\ >0& \text{ for all }x\in ℝ\setminus \{0\}\end{array}$

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

The fact that $g$ with $g(x)=x^3$ is strictly' monotonously increasing, although only $f^{\prime }\ge 0$ and not $f^{\prime }>0$, 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 monotonous, even if there is not everywhere $f^{\prime }>0$.

Question: Why is $g:ℝ\to ℝ:x\mapsto x^3$ strictly monotonously increasing on $ℝ$?

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

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

So $g$ is strictly monotonously increasing on all of $ℝ$ .

Warning

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

Exponential and logarithm function

Example (Monotony of the exponential and logarithm function)

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

$f^{\prime }(x)=\exp (x)>0$

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

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

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

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

There is

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

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

Trigonometric functions

Example (Monotony of the sine function)

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

$f^{\prime }(x)=\cos (x)\{\begin{array}{cc}>0& \text{ for }x\in (-\frac{\pi }{2}+2\pi k,\frac{\pi }{2}+2\pi k),k\in \mathbb{Z},\\ <0& \text{ for }x\in (\frac{\pi }{2}+2\pi k,\frac{3\pi }{2}+2\pi k),k\in \mathbb{Z}\end{array}$

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

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

Here, $g^{\prime }(x)=-\sin (x)\{\begin{array}{cc}>0& \text{ for }x\in (\pi +2\pi k,2\pi +2\pi k),k\in \mathbb{Z},\\ <0& \text{ for }x\in (2\pi k,\frac{3\pi }{2}+\pi +2\pi k),k\in \mathbb{Z}\end{array}$.

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

Example (Monotony of the tangent function)

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

${\tan }^{\prime }(x)=1+\underset{\ge 0}{\underbrace{{\tan }^2(x)}}\ge 1>0$

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

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

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

${\cot }^{\prime }(x)=-1-\underset{\ge 0}{\underbrace{{\tan }^2(x)}}\le -1<0$

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

Exercise

Monotony intervals and existence of a zero

Exercise (Monotony intervals and existence of a zero)

Where is the following polynomial function monotonous?

$f:ℝ\to ℝ,f(x)=x^3-x-1$

Prove that $f$ has exactly one zero.

Solution (Monotony intervals and existence of a zero)

Graph of the function $f(x)=x^3-x-1$

Monotony intervals:

The function $f$ is differentiable on all of $ℝ$ , with

$f^{\prime }(x)=3x^2-1=3(x^2-\frac{1}{3})=3(x+\frac{1}{\sqrt{3}})(x-\frac{1}{\sqrt{3}})$

So

$f^{\prime }(x)>0⟺\{\begin{array}{c}x+\frac{1}{\sqrt{3}}>0\text{ and }x-\frac{1}{\sqrt{3}}>0⟺x>-\frac{1}{\sqrt{3}}\text{ and }x>\frac{1}{\sqrt{3}}⟺x>\frac{1}{\sqrt{3}}\\ x+\frac{1}{\sqrt{3}}<0\text{ and }x-\frac{1}{\sqrt{3}}<0⟺x<-\frac{1}{\sqrt{3}}\text{ and }x<\frac{1}{\sqrt{3}}⟺x<-\frac{1}{\sqrt{3}}\end{array}$

According to the monotony criterion, $f$ is strictly monotonously increasing on $(-\mathrm{\infty },-\frac{1}{\sqrt{3}})$ and on $(\frac{1}{\sqrt{3}},\mathrm{\infty })$ . Further,

$f^{\prime }(x)<0⟺\{\begin{array}{c}x+\frac{1}{\sqrt{3}}>0\text{ and }x-\frac{1}{\sqrt{3}}<0⟺x>-\frac{1}{\sqrt{3}}\text{ and }x<\frac{1}{\sqrt{3}}⟺-\frac{1}{\sqrt{3}}<x<\frac{1}{\sqrt{3}}\\ x+\frac{1}{\sqrt{3}}<0\text{ and }x-\frac{1}{\sqrt{3}}>0⟺x<-\frac{1}{\sqrt{3}}\text{ and }x>\frac{1}{\sqrt{3}}⟺\text{ not possible}\end{array}$

According to the monotony criterion, $f$ is strictly monotonously decreasing on $(-\frac{1}{\sqrt{3}},\frac{1}{\sqrt{3}})$ .

$f$ has exactly one zero:

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

$\begin{array}{ccccccc}x& -1& -\frac{1}{\sqrt{3}}& 0& \frac{1}{\sqrt{3}}& 1& 2\\ f(x)& -1& -1+\frac{2}{3\sqrt{3}}& -1& -1-\frac{2}{3\sqrt{3}}& -1& 5\end{array}$

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

• On $(-\mathrm{\infty },-\frac{1}{\sqrt{3}})$ is $f$ strictly monotonously increasing. Because of $f(-\frac{1}{\sqrt{3}})=-1+\frac{2}{3\sqrt{3}}<0$ there is $f(x)<0$ for all $x\in (-\mathrm{\infty },-\frac{1}{\sqrt{3}})$.
• On $(-\frac{1}{\sqrt{3}},\frac{1}{\sqrt{3}})$ is $f$ then strictly monotonously decreasing. So there is also $f(x)<0$ for all $x\in (-\frac{1}{\sqrt{3}},\frac{1}{\sqrt{3}})$.
• Subsequently $f$ increases on $(\frac{1}{\sqrt{3}},\mathrm{\infty })$ again strictly monotonously. 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 monotony of $f$ on $(\frac{1}{\sqrt{3}},\mathrm{\infty })$ , there cannot be any further zeros.

Necessary and sufficient criterion for strict monotony

Exercise (Necessary and sufficient criterion for strict monotony)

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

1. $f^{\prime }(x)\ge 0$ for all $x\in (a,b)$
2. The zero set of $f^{\prime }$ contains no open interval.

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

Proof (Necessary and sufficient criterion for strict monotony)

From the monotony criterion we already know that $f$ is monotonously increasing exactly when $f^{\prime }(x)\ge 0$. So we only have to show that $f$ is strictly monotonously increasing exactly when the second condition is additionally fulfilled.

$⟹$: $f$ strictly monotonously increasing $⟹$ the set of zeros of $f^{\prime }$ does not contain an open interval.

We perform a proof by contradiction. In other words, we show: If the set of zeros of $f^{\prime }$ contains an open interval, $f$ is not strictly monotonously increasing. Assume there is $a_1,b_1\in (a,b)$ with $f^{\prime }(x)=0$ for all $x\in (a_1,b_1)$. Then, by the mean value theorem there is a $\xi \in (a_1,b_1)$ with

$f(b_1)-f(a_1)=\underset{=0}{\underbrace{f^{\prime }(\xi )}}(b_1-a_1)=0$

So $f(a_1)=f(b_1)$. If now $x\in (a_1,b_1)⟺a_1<x<b_1$, then since $f$ is monotonously increasing, there is

$f(a_1)\le f(x)\le f(b_1)$

So there is $f(x)=f(b_1)$ for all $x\in (a_1,b_1)$. Hence, $f$ is not strictly monotonously increasing.

$⟸$: the set of zeros of $f^{\prime }$ does not contain an open interval $⟹$ $f$ strictly monotonously increasing

We perform a proof by contradiction. In other words, we show: if $f$ is monotonously, but not strictly monotonously increasing, then the zero set of $f^{\prime }$ 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 monotony of $f$ there is

$f(a_2)\le f(x)\le f(b_2)$

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 $x\in (a_2,b_2)$:

$f^{\prime }(x)=0$

so the set of zeros of $f^{\prime }$ does contain an open interval and we get a contradiction.

Exercise: $f:ℝ\to ℝ,f(x)=x-\sin (x)$ is strictly monotonously increasing

$f$ is differentiable for all $x\in ℝ$ where

$f^{\prime }(x)=1-\cos (x)\ge 0$

as $\cos (x)\le 1$ for all $x\in ℝ$. Hence $f$ is monotonously increasing. Further there is

$\begin{array}{cc}& f^{\prime }(x)=1-\cos (x)=0\\ ⟺& \cos (x)=1\\ ⟺& x\in \{2\pi k\mid k\in \mathbb{Z}\}\end{array}$

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