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Letting $f(x)=2xe^x-1$

$f(0)=-1<0$

$f(1)=2e-1>0$

So by the IVT there must be a solution in the interval $(0,1)$. But I do not understand how I can show the uniqueness of a solution.

Disintegrating By Parts
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AlisonS
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1 Answers1

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Hint: Check the derivative of $f(x)=2xe^x-1$ on $[0,1]$ to see that $f$ is strictly increasing on that interval, which will allow you to conclude uniqueness (why?).

csch2
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  • So f’(x)=2e^x+2xe^x=2e^x (1+x)=0 The derivative changes its sign at the point -1 That means f(x) is strictly increasing on that interval (0,1). Strictly increasing functions have to be continuous. But still do not understand how this will help us to prove uniqueness. – AlisonS Apr 29 '20 at 21:10
  • If $f(x)=f(y)=0$ for some $x,y\in[0,1]$, then by Rolle's theorem $\exists c\in(x,y)$ such that $f'(c)=0$. Can you see why this is a contradiction? – csch2 Apr 29 '20 at 21:12
  • f is strictly increasing on the interval (0,1) that means there can’t be f(x)=f(y) for any x,y. – AlisonS Apr 29 '20 at 21:25
  • Exactly. So any solution has to be unique, otherwise you would have $f(x)=f(y)=0$ for two different $x,y$, which would contradict Rolle's theorem. – csch2 Apr 29 '20 at 21:32
  • thank you very much! – AlisonS Apr 29 '20 at 21:34