Monotone functions – Serlo

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 $f'$ 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 $f'$ 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 $f'(x)\geq 0$ and if the function is monotonously 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 monotonous, but $f'(0)=3\cdot 0^{2}=0$.

Theorem (Monotony criterion for differentiable functions)

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

1. $f'\geq 0$ on $(a,b)$ $\iff$ $f$ monotonously increasing on $[a,b]$
2. $f'\leq 0$ on $(a,b)$ $\iff$ $f$ monotonously decreasing on $[a,b]$
3. $f'>0$ on $(a,b)$ $\Longrightarrow$ $f$ strictly monotonously increasing on $[a,b]$
4. $f'<0$ on $(a,b)$ $\Longrightarrow$ $f$ strictly monotonously 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 (monotony 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 monotonously 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 monotonously increasing.

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

3. $\Longrightarrow$: $f'>0$ 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 $f(x_{1})\geq f(x_{2})$. 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 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 $f(x_{1})\leq f(x_{2})$. 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 monotonously 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 monotony, $f(x_{1})\leq f(x_{2})$. 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 monotony 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 monotonously 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 monotony, $f(x_{1})\geq f(x_{2})$. 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: monotony criterion

Quadratic and cubic functions

Example (Monotony of quadratic and cubic functions)

Graphs of the functions $f$ and $g$

Graphs of the functions $f'$ and $g'$

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

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

Question: Why is $g:\mathbb {R} \to \mathbb {R} :x\mapsto x^{3}$ strictly monotonously 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 monotony criterion. So we only have to look at the case $x<0<y$. Here there is with the arrangement axioms (missing):

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

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

Warning

In the example $x\mapsto x^{3}$ we have seen that the statement "$f'>0$ 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 $f'>0$. 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 monotonous falling" does not hold true in general.

Exponential and logarithm function

Example (Monotony of the exponential and logarithm function)

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

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

Therefore, according to the monotony criterion, $\exp$ is strictly monotonously 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 monotonously 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 monotonously 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 monotonously decreasing on $\mathbb {R} ^{-}$.

Trigonometric functions

Example (Monotony 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 monotonously increasing on the intervals $[-{\tfrac {\pi }{2}}+2\pi k,{\tfrac {\pi }{2}}+2\pi k]$ and strictly monotonously 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 monotonous 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 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 $\tan :D=\mathbb {R} \setminus \{{\tfrac {\pi }{2}}+k\pi \mid k\in \mathbb {Z} \}\to \mathbb {R} ,\ \tan(x)={\tfrac {\sin(x)}{\cos(x)}}$ 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 monotonously 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 monotonous 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 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?

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

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 $\mathbb {R}$ , 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 monotony criterion, $f$ is strictly monotonously 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 monotony criterion, $f$ is strictly monotonously 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 monotonously 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 monotonously 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 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 $\left({\tfrac {1}{\sqrt {3}}},\infty \right)$ , 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 $f:[a,b]\to \mathbb {R}$ which is differentiable to $(a,b)$ is strictly monotonously increasing exactly when there is

1. $f'(x)\geq 0$ for all $x\in (a,b)$
2. 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 monotonously increasing on all of $\mathbb {R}$ .

Proof (Necessary and sufficient criterion for strict monotony)

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

$\Longrightarrow$: $f$ strictly monotonously 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 monotonously 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 monotonously 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 monotonously increasing.

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

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

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 monotonously 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 monotonously 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 monotonously increasing on $\mathbb {R}$ .