Increasing and decreasing interval of this function |e^x+e^{-x}|

[WMDhamnekar]

Hello,

I want to know what is the incresing and decreasing interval of this even function $|e^x+e^{-x}|?$

If any member knows the correct answer, may reply to this question.

 

[MarkFL]

Since \(0<e^{x}+e^{-x}\) for all real \(x\), we may simply write:

$$f(x)=e^{x}+e^{-x}$$

You've posted this question in our Pre-Calculus forum, so I am assuming you wish not to utilize differential calculus in the analysis of this function's behavior. Is this correct?

 

[WMDhamnekar]

Since \(0<e^{x}+e^{-x}\) for all real \(x\), we may simply write:

$$f(x)=e^{x}+e^{-x}$$

You've posted this question in our Pre-Calculus forum, so I am assuming you wish not to utilize differential calculus in the analysis of this function's behavior. Is this correct?

Hello,
I want to determine whether this sequence $M_n^{(1)}=\frac{e^{\theta*S_n}}{(\cosh{\theta})^n} \tag{1}$ is martingale.

For checking the integrability of (1), I want to know on which interval $|e^x+e^{-x}|$ is increasing and decreasing. What is the supremum of this even function? One math expert informed me online that it is increasing on $(-\infty,0)$and decreasing on $(0, \infty)$and its supremum is $2^{-n}$ at $x or \theta=0$ How?

If you think this question doesn't belong to "Precalculus" forum, You may move it to "Advanced probability and statictics" or any other forum, you may deem fit :)

 

[MarkFL]

I have moved the thread as per your suggestion.

With regards to your original question, let's go back to:

$$f(x)=e^{x}+e^{-x}$$

We find:

$$f'(x)=e^x-e^{-x}$$

Equating this to zero, there results:

$$e^{2x}=1$$

Which implies:

$$x=0$$

So, we know the function has 1 turning point, at \((0,2)\). We observe that:

$$f''(x)=f(x)$$

And:

$$f''(0)=f(0)=2>0$$

This tells us the function is concave up at the turning point, which thus implies this turning point is a minimum, and is in fact the global minimum. Hence the function is decreasing on:

$$(-\infty,0)$$

And increasing on:

$$(0,\infty)$$