# Monotone functions – Serlo

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## 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 ${\displaystyle f'}$ of a differentiable function ${\displaystyle f}$ is non-negative or (non-positive) on an interval ${\displaystyle (a,b)}$ , then ${\displaystyle f}$ is monotonously increasing (or decreasing) on ${\displaystyle (a,b)}$. If ${\displaystyle f'}$ is even strictly positive (or negative) ${\displaystyle (a,b)}$, then ${\displaystyle f}$ 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 ${\displaystyle [a,b]}$, then ${\displaystyle f'(x)\geq 0}$ and if the function is monotonously decreasing on ${\displaystyle [a,b]}$, then then ${\displaystyle f'(x)\geq 0}$. However, the inversion does not hold true in the strict case, monotone functions do not always have ${\displaystyle f'(x)>0}$ or ${\displaystyle f'(x)<0}$ . For instance, ${\displaystyle f(x)=x^{3}}$ is strictly monotonous, but ${\displaystyle f'(0)=3\cdot 0^{2}=0}$.

Theorem (Monotony criterion for differentiable functions)

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

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

## Proof

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

Proof (monotony criterion for differentiable functions)

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

Now, the two directions "${\displaystyle \Longleftarrow }$" follow:

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

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

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

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

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

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

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

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

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

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

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

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

## Examples: monotony criterion

Example (Monotony of quadratic and cubic functions)

Graphs of the functions ${\displaystyle f}$ and ${\displaystyle g}$
Graphs of the functions ${\displaystyle f'}$ and ${\displaystyle g'}$

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

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

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

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

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

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

The fact that ${\displaystyle g}$ with ${\displaystyle g(x)=x^{3}}$ is strictly' monotonously increasing, although only ${\displaystyle f'\geq 0}$ and not ${\displaystyle f'>0}$, 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 ${\displaystyle f'>0}$.

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

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

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

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

Warning

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

${\displaystyle f'(x)=\exp(x)>0}$

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

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

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

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

There is

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

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

### Trigonometric functions

Example (Monotony of the sine function)

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

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

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

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

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

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

Example (Monotony of the tangent function)

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

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

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

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

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

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

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

## Exercise

### Monotony intervals and existence of a zero

Exercise (Monotony intervals and existence of a zero)

Where is the following polynomial function monotonous?

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

Prove that ${\displaystyle f}$ has exactly one zero.

Solution (Monotony intervals and existence of a zero)

Graph of the function ${\displaystyle f(x)=x^{3}-x-1}$

Monotony intervals:

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

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

So

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

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

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

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

${\displaystyle f}$ has exactly one zero:

For ${\displaystyle f}$ , we have the following table of values:

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

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

• On ${\displaystyle \left(-\infty ,-{\tfrac {1}{\sqrt {3}}}\right)}$ is ${\displaystyle f}$ strictly monotonously increasing. Because of ${\displaystyle f\left(-{\tfrac {1}{\sqrt {3}}}\right)=-1+{\tfrac {2}{3{\sqrt {3}}}}<0}$ there is ${\displaystyle f(x)<0}$ for all ${\displaystyle x\in \left(-\infty ,-{\tfrac {1}{\sqrt {3}}}\right)}$.
• On ${\displaystyle \left(-{\tfrac {1}{\sqrt {3}}},{\tfrac {1}{\sqrt {3}}}\right)}$ is ${\displaystyle f}$ then strictly monotonously decreasing. So there is also ${\displaystyle f(x)<0}$ for all ${\displaystyle x\in \left(-{\tfrac {1}{\sqrt {3}}},{\tfrac {1}{\sqrt {3}}}\right)}$.
• Subsequently ${\displaystyle f}$ increases on ${\displaystyle \left({\tfrac {1}{\sqrt {3}}},\infty \right)}$ again strictly monotonously. Because of ${\displaystyle f(1)=-1<0}$ and ${\displaystyle f(2)=5>0}$, there must be an ${\displaystyle x_{0}\in (1,2)}$ with ${\displaystyle f(x_{0})=0}$ by the mean value theorem. Because of the strict monotony of ${\displaystyle f}$ on ${\displaystyle \left({\tfrac {1}{\sqrt {3}}},\infty \right)}$ , 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 ${\displaystyle f:[a,b]\to \mathbb {R} }$ which is differentiable to ${\displaystyle (a,b)}$ is strictly monotonously increasing exactly when there is

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

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

Proof (Necessary and sufficient criterion for strict monotony)

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

${\displaystyle \Longrightarrow }$: ${\displaystyle f}$ strictly monotonously increasing ${\displaystyle \Longrightarrow }$ the set of zeros of ${\displaystyle f'}$ does not contain an open interval.

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

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

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

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

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

${\displaystyle \Longleftarrow }$: the set of zeros of ${\displaystyle f'}$ does not contain an open interval ${\displaystyle \Longrightarrow }$ ${\displaystyle f}$ strictly monotonously increasing

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

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

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

${\displaystyle f'(x)=0}$

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

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

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

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

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

{\displaystyle {\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 ${\displaystyle f'}$ contains only isolated points, and thus no open interval. Therefore ${\displaystyle f}$ is strictly monotonously increasing on ${\displaystyle \mathbb {R} }$ .