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I want to check the monotonicity of the function for $x>0$ $$\frac{\cosh 2 x^3 }{3\cosh 5 x^3 }$$ Computing the first derivative, it can be proved that it is negative and then the function is decreasing.

My question is can we claim that since $\,\cosh x^3\,$ is an increasing function for $x>0$, and since the numerator is less than the denominator, then, the function is decreasing?

charmin
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  • It's not clear what you're asking. What facts are you saying you already know, and what statement are you trying to deduce from it? In your first paragraph, "increasing" seems to contradict the first derivative being positive. – Greg Martin Sep 14 '20 at 16:22
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    See the alternate form from this link https://www.wolframalpha.com/input/?i=derivative+ln%28cosh%282x%5E3%29%2F%283cosh%285x%5E3%29%29%29 . – Miss and Mister cassoulet char Sep 14 '20 at 17:22
  • @GregMartin It was a typo. I meant decreasing! – charmin Sep 14 '20 at 19:16
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    Your question seems to be asking whether this implication is true: if $f(x)$ is increasing and $f(x)<g(x)$, then $f(x)/g(x)$ is decreasing. This is definitely not true, as examples like $f(x)=1-1/x$, $g(x)=1$ show. – Greg Martin Sep 14 '20 at 20:17
  • @GregMartin Thanks. In your example, $f$ and $g$ are not of the same type. So, in the general case, it is not true. – charmin Sep 14 '20 at 21:40
  • But what does "same type" mean, exactly? Anyway, in general it's better to have these statements given precisely—that makes it easier to evaluate whether they're true or false. – Greg Martin Sep 14 '20 at 22:55

1 Answers1

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We have that by $y=x^3>0$

$$\frac{\cosh 2 x^3 }{3\cosh 5 x^3 }=\frac{\cosh 2y }{3\cosh 5 y }=\frac{e^{3y}(e^{4y}+1)}{3(e^{10y}+1)}$$

and $e^y=z>1$

$$\frac{e^{3y}(e^{4y}+1)}{e^{10y}+1}=\frac{z^3(z^4+1)}{z^{10}+1}=f(z)$$

and

$$f'(z)= \frac{z^2(-3z^{14}-7z^{10}+7z^4+3)}{(z^{10}+1)^2}$$

and

$$-3z^{14}-7z^{10}+7z^4+3=-3(z^{14}-1)-7z^4(z^6-1)<0$$

user
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  • So, my claim is not true and we need to prove it. Thanks. – charmin Sep 14 '20 at 19:20
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    @charmin Yes indeed by your claim we can conclude that $f(x)<1$ but for monotonicity we need to prove that $f'(x)<0$. The way I've shown should simplify this evaluation. – user Sep 14 '20 at 19:29
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    @charmin I can't exclude there is sime other way to get the same conclusion in this partcular case. I'll take a look also on that. You are welcome! Bye – user Sep 14 '20 at 19:30
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    @charmin Refer to the related OP. – user Sep 14 '20 at 19:34