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

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- MHB
- Thread starter WMDhamnekar
- Start date

The supremum of the function is then f(0)=2.In summary, the function \(|e^x+e^{-x}|\) is decreasing on \((-\infty,0)\), increasing on \((0,\infty)\), and has a supremum of 2 at \(x=0\). This information can be used to determine the increasing and decreasing intervals of the even function \(|e^x+e^{-x}|\) in the given sequence.

- #1

- 375

- 28

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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- #2

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MHB

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\(\displaystyle f(x)=e^{x}+e^{-x}\)

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?

- #3

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Hello,MarkFL said:

\(\displaystyle f(x)=e^{x}+e^{-x}\)

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?

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 :)

- #4

Gold Member

MHB

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

\(\displaystyle f(x)=e^{x}+e^{-x}\)

We find:

\(\displaystyle f'(x)=e^x-e^{-x}\)

Equating this to zero, there results:

\(\displaystyle e^{2x}=1\)

Which implies:

\(\displaystyle x=0\)

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

\(\displaystyle f''(x)=f(x)\)

And:

\(\displaystyle f''(0)=f(0)=2>0\)

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:

\(\displaystyle (-\infty,0)\)

And increasing on:

\(\displaystyle (0,\infty)\)

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