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# Differentiation

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Hi I have tried to tackle the following problem :

My solution is as follows:but it is wrong: the official answer shows the power of x as n/(n-1)  (see end)

Thanks for your time and attention...best regards & keep safe

Feb 17, 2021

### 8+0 Answers

#1
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Hi OldTimer, it is nice to see you again.

I got the same as you until the end bit.

For stationary points

$$\frac{2}{n}x^{(\frac{2}{n}-1)}=\frac{n+1}{n}x^{(\frac{n+1}{n}-1) }\\ \frac{2}{n}x^{(\frac{2}{n}-1)}=\frac{n+1}{n}x^{(\frac{1}{n}) }\\ 2x^{(\frac{2}{n}-1)}=(n+1)x^{(\frac{1}{n}) }\\ 2x^{(\frac{2}{n}-1-\frac{1}{n})}=(n+1) \\ x^{(\frac{1-n}{n})}=\frac{n+1}{2} \\ x=\left(\frac{n+1}{2}\right)^{\frac{n}{1-n}} \\$$

You should do more to show it is a maximum and not some other kind of stationary point though.

This desmos graph shows that it is correct:

https://www.desmos.com/calculator/8hkpyrldi4

LaTex:

\frac{2}{n}x^{(\frac{2}{n}-1)}=\frac{n+1}{n}x^{(\frac{n+1}{n}-1)  }\\
\frac{2}{n}x^{(\frac{2}{n}-1)}=\frac{n+1}{n}x^{(\frac{1}{n})  }\\
2x^{(\frac{2}{n}-1)}=(n+1)x^{(\frac{1}{n})  }\\
2x^{(\frac{2}{n}-1-\frac{1}{n})}=(n+1)  \\
x^{(\frac{1-n}{n})}=\frac{n+1}{2}  \\
x=\left(\frac{n+1}{2}\right)^{\frac{n}{1-n}} \\

Feb 17, 2021
#2
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Excellent, Melody   !!!!

CPhill  Feb 17, 2021
#3
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Thanks Chris

To be truthful, I have not finished it.

I am not sure how easy or difficult it would be to show it is a maximum.

Melody  Feb 17, 2021
edited by Melody  Feb 17, 2021
#4
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Hi Melody....thanks for your prompt  reply.

This is the answer (no 10)  offered in the book which also differs from your solution......heck (2nd time I met this circumstance)!

re 'do more to show it is a Max value' : I extracted 2 nd derivative which is a negative...thus pointing to a maximum value.

Thanks for your feedback & knowledge!

Feb 18, 2021
#5
+113779
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It is not different from my answer, they are exactly the same.  I'll show you.

$$x=\left(\frac{n+1}{2}\right)^{\frac{n}{1-n}} \\~\\ x=\left(\frac{n+1}{2}\right)^{\frac{-n}{n-1}} \\~\\ x=\left(\frac{2}{n+1}\right)^{\frac{n}{n-1}} \\~\\$$

And you are always welcome :)

Melody  Feb 18, 2021
#6
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Hi Melody...so I had another look at the question (thanks to your feedback) and tried to keep it simple.

So could the question be saying that the derivative of main function  is a maximum and is equal to square root of same function (ie  +/- values apply)...and therefore this is given and no need to prove it??!!

In that case I get the following for x (Word is terrible for writing power/equations so I hope there are no typos) - which agrees with book. Regards & thanks again

Feb 18, 2021
#7
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Mine agreed with your book too.

I am a bit confused.  Do you have a question?

I thought you already showed that it was a maximum.

Melody  Feb 18, 2021
edited by Melody  Feb 18, 2021
edited by Melody  Feb 18, 2021
#8
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Hi Melody...apolgies for confusing you.....thanks for your patience in re-clarifying...such a great help...you are really talented!!

OldTimer  Feb 19, 2021