# Mean value theorem - prove inequality

• I
• physics1000
I see, the function is increasing from negative over ##f(1)=0## to positive.In summary, the student is trying to find a function that is increasing from negative over ##f(1)=0## to positive. However, they get lost and make a mistake.f

#### physics1000

TL;DR Summary
##(x+1)\ln(x)>2(x-1)##
I tried using mean value theorem with ##f(x)=\ln(x)##, used the points: ##[1,x]##, didn't manage to get it. Then thought of using Cauchy value theorem with ##f(x)=\ln(x)## and ##g(x)=(x-1)/(x+1)## or something like that, tried other stuff, nothing helped...
I don't need an answer (although I don't have sadly, it's from a test).
I need just a tip on how to start it...
i cannot use Taylor in here (##\ln(x)## is not Taylor function), therefore, its only MVT, but I don't know which point I should try... since I must get the annoying ##\ln(x)## thingy...
The only thing I can think which is smart, I must do MVT twice, one with ##[1,x+1],## show its bigger than something, and then go back to original or something like that... any tip will be welcomed.

I corrected your TeX code. And did you forget to mention that ##x>1?##

Here is explained how you can type formulas on PF: https://www.physicsforums.com/help/latexhelp/

*** mentor note: changed .om to .com url should now work.

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• physics1000
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Actually the link won't work. it shows me
:

### This site can’t be reached​

www.physicsforums.com’s server IP address could not be found.

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Differentiation looks promising from a look at the function ##f(x)=(x+1) \log(x) - 2(x-1).##

Differentiation looks promising from a look at the function ##f(x)=(x+1) \log(x) - 2(x-1).##
Oh, I never thought of it... it seems a little weird it will work, but I will give it a try and update :)

Oh its the same, just # instead of $Alright :) Here is explained how you can type formulas on PF: https://www.physicsforums.com/help/latexhelp/ Oh its the same, just # instead of$
Alright :)
Double # for inline code and double \$ for extra lines. And a \ in front of ln.

I lost the c in com somehow (link).

• physics1000
Don't forget post #5.

• physics1000
Differentiation looks promising from a look at the function ##f(x)=(x+1) \log(x) - 2(x-1).##
Oh i actually managed to solve it, thanks
but just something so I know.
how am I supposed to know that:
##x^2ln(x)-x+1>x-1##, when I did the derivative, I used calculator to think of it, that was the only way it should work, if its bigger then 1, so i just did some numbers and saw it.
How am i supposed to know its bigger then 1 without calculator?

The function ##f(x)=(x+1) \log(x) - 2(x-1)## is zero at ##x=1## and the derivative is ##f'(x)=-1+\ln(x)+1/x.## This is zero at ##x=1## and positive everywhere else. Thus ##f(x)## is increasing everywhere: from negative over ##f(1)=0## to positive.

The function ##f(x)=(x+1) \log(x) - 2(x-1)## is zero at ##x=1## and the derivative is ##f'(x)=-1+\ln(x)+1/x.## This is zero at ##x=1## and positive everywhere else. Thus ##f(x)## is increasing everywhere: from negative over ##f(1)=0## to positive.
Oh you probably sorted the derivative better then me if you got that, I got something way more complicated lol

##xln\left(x\right)+\frac{x+1}{x}-2## this is what I got.. after some sorting
##\frac{x^2ln\left(x\right)-x+1}{x-1}>1##
but you understand? i can't know its bigger then 1 like that

I used the internet ... Let's check:

\begin{align*}
[(x+1) \log(x) - 2(x-1)]'&=(x+1)'\log x + (x+1)\left[\log(x)\right]'-2\\
&=\log(x) +\left(\dfrac{x+1}{x}\right)-2\\
&=\log(x) + \dfrac{x}{x}+\dfrac{1}{x} -2\\
&=\log(x) -1 +\dfrac{1}{x}
\end{align*}

I used the internet ... Let's check:

\begin{align*}
[(x+1) \log(x) - 2(x-1)]'&=(x+1)'\log x + (x+1)\left[\log(x)\right]'-2\\
&=\log(x) +\left(\dfrac{x+1}{x}\right)-2\\
&=\log(x) + \dfrac{x}{x}+\dfrac{1}{x} -2\\
&=\log(x) -1 +\dfrac{1}{x}
\end{align*}
ohh, i had a mistake lol, i forgot to remove the ##X## before the ##ln(x)##... so from ##x>1## it is positive.. I see, but the problem is, I chose ##[1,x]##, or it doesn't matter?

Ahh, actually i understand, but I am really having trouble thinking about that function.. I would have never thought about it..
it isn't possible using ##f(x)=ln(x)## I guess, right?

To be honest: I wanted to know whether the formula is right, especially if you had forgotten to mention ##x>1## or not. I wasn't sure what happens between ##0## and ##1## in particular what happens left and right of ##1/e.## But I'm lazy, and to feed WolframAlpha with both sides wouldn't tell me enough, so I simply typed
https://www.wolframalpha.com/input?i=y=(x+1)+\log(x)+-+2(x-1)
and the graph told me the rest.

• physics1000
To be honest: I wanted to know whether the formula is right, especially if you had forgotten to mention ##x>1## or not. I wasn't sure what happens between ##0## and ##1## in particular what happens left and right of ##1/e.## But I'm lazy, and to feed WolframAlpha with both sides wouldn't tell me enough, so I simply typed
https://www.wolframalpha.com/input?i=y=(x+1)+\log(x)+-+2(x-1)
and the graph told me the rest.
Oh dam, sorry! I really forgot about ##X>1##!, didnt notice it!
Thanks :) I will learn from it!

One useful principle is the fact that a smooth function cannot change direction except where the derivative equals zero. So to investigate the function ln(x) - [2(x-1)/(x+1)^2], one can take the derivative and see it only equals zero at x=1. Then it helps to look at the limiting behavior of the function at the end points of the domain. but the limit as x-->0+ is minus infinity, so this function is surely not always positive. On the other hand the limit is plus infinity, as x-->infinity, so it never changes direction.
To find where it is positive then one sets it equal to zero, and it will be positive precisely to the right of that point. it might not be so easy to find that point unless one notes that x=1 works, or unless one is given that this is the interesting point. It helps that this the only zero of the derivative so one would naturally also look at the value there.
In general one always looks at the (possibly limiting) behavior at end points, as well as zeroes of the derivative, which are the only places where the direction of the graph can change.