Proving that a Function does not have any extreme points

Question

I need to find a continuous function, such that $$\lim_{x \to \infty}f(x)=L_{1}$$ $$\lim_{x \to -\infty}f(x)=L_{2}$$

And such that $f$ doesn't have any extreme points.

So I found this function here: $f(x)=\frac{x}{|x|+1}$ for every $x \in \mathbb{R}$.

I have proved that the function is continuous and also that $\lim_{x \to \infty}f(x)=1$ and $\lim_{x \to -\infty}f(x)=-1$.

Now all I left to do is proving that this function has no extreme points at all.

I want to prove that the function is strictly increasing, and for that reason my theorem holds.

I have tried to choose $x_{1}, x_{2} \in \mathbb{R}$ such that $x_{1}<x_{1}$ and prove that $f(x_{1})<f(x_{2})$ but I find this difficult for me for some reason.

I will appreciate you help to finish my proof!

Thank you so much! :)

Answer 1

Simply prove $-1 < \frac {x}{|x| + 1} < 1$ for all $x$.

Thus for any possible $x_0$ we have $-1 < f(x_0) < 1$.

And as $\lim_{x\to \infty} = 1$ we have $x$s where $f(x_0) < f(x) < 1$.

And as $\lim_{x\to -\infty} = -1$ we havve $x$s where $-1 < f(x) < f(x_0)$.

So $f(x_0)$ can not be an extreme value for any $x_0$.

....

If you want to show $f$ is increasing do it in cases

Cases $x < y < 0$, $x < 0 \le y$, $0\le x < y$.

Note that $\frac {|w|}{|w| + 1} = \frac {|w|+1}{|w|+1} - \frac 1{|w|}=1 -\frac 1{|w|}$

If $0 \le x < y$ then $x=|x|;y=|y|$ and $\frac 1{|y|} < \frac 1{|x|}$ and so $\frac {x}{|x| + 1} =\frac {|x|}{|x|+1} = 1 -\frac 1{|x|} < 1 -\frac 1{|y|} = \frac {y}{|y| + 1}$.

If $x < 0 \le y$ then $\frac {x}{|x|+1} < 0 \le \frac {y}{|x| + 1}$.

And if $x < y < 0$ then $|x| > |y|$ and

$\frac {x}{|x| + 1} = -\frac {|x|}{|x|+1} = -(1-\frac 1{|x|})< -(1-\frac 1{|y|}) = \frac y{|y| + 1}$.

Answer 2

One we is to prove the function to be derivable for x $\in$ R. Then show derivative > 0.

Answer 3

Consider the function $f\left(x\right)=\frac{e^{x}}{1+e^{x}}$

$f$ is continuous by arithmetic property of continuous functions. Also We have:

$$\underset{x\rightarrow\infty}{\lim}\frac{e^{x}}{1+e^{x}}=1$$

$$\underset{x\rightarrow-\infty}{\lim}\frac{e^{x}}{1+e^{x}}=0$$

$f$ doesn't have a max because for every $x_1\in\mathbb{R}$ we can take $x_1+1$ such that:

$$\underbrace{f\left(x_1+1\right)}=\frac{e^{x_{1}+1}}{1+e^{x_{1}+1}}=\frac{e\cdot e^{x_{1}}}{1+e\cdot e^{x_{1}}}=\frac{e^{x_{1}}}{\frac{1}{e}+e^{x_{1}}}\underset{\frac{1}{e}<1}{\underbrace >}\frac{e^{x_{1}}}{1+e^{x_{1}}}=\underbrace{f\left(x_{1}\right)}$$

Also, $f$ doesn't have a min because for every $x_2\in\mathbb{R}$ we can take $x_2-1$ such that:

$$\underbrace{f\left(x_2-1\right)}=\frac{e^{x_{2}-1}}{1+e^{x_{2}-1}}=\frac{\frac{e^{x_{2}}}{e}}{1+\frac{e^{x_{2}}}{e}}=\frac{e^{x_{2}}}{e+e^{x_{2}}}\underset{e>1}{\underbrace <}\frac{e^{x_{2}}}{1+e^{x_{2}}}=\underbrace{f\left(x_{2}\right)}$$