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

AI Thread Summary
The function |e^x + e^{-x}| is analyzed for its increasing and decreasing intervals. It is established that the function is decreasing on the interval (-∞, 0) and increasing on (0, ∞), with a global minimum at the turning point (0, 2). The function's behavior is confirmed through the first and second derivatives, indicating concavity and the nature of the turning point. The supremum of the function occurs at x = 0, where it reaches the value of 2. This analysis provides a clear understanding of the function's intervals of increase and decrease.
WMDhamnekar
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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.
 
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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?
 
MarkFL said:
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?:confused:

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 :)
 
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:

$$(0,\infty)$$
 
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