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The equation $$e^{x^3-x}-2$$

  1. has no solution in $[0,1]$
  2. has a unique solution in $[0,1]$
  3. has two solutions in $[0,1]$
  4. has four solutions in $[0,1]$

Correct answer is B but how do you approach a problem like this? Can you post a solution please? What does $[0,1]$ even mean?

Thank you

(thanks for the edit)

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    $\begingroup$ I assumed you meant $e^{x^3}-x-2$. If you instead meant $e^{x^3-x}-2$ or $e^{x^3-x-2}$, please change it to the correct one by clicking the "edit" link on the bottom of your post. $\endgroup$ – Arthur Jan 18 '18 at 6:43
  • $\begingroup$ its all good thank you :) $\endgroup$ – Deni Katsman Jan 18 '18 at 6:43
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    $\begingroup$ Also "in [0,1]" means "between $0$ and $1$ (including $0$ and $1$)". $\endgroup$ – Arthur Jan 18 '18 at 6:45
  • $\begingroup$ Fix your title please $\endgroup$ – Mohammad Zuhair Khan Jan 18 '18 at 6:47
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    $\begingroup$ @XcoderX Your edit overwrote and undid OP's previous edit (and then the title and the body of the question no longer matched). This is why I rolled back your edit. Let the OP decide what they really meant to ask. $\endgroup$ – dxiv Jan 18 '18 at 6:49
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you have asked " What does $[0, 1]$ mean?"

It means all $x$ real numbers $x$ such that $ 0\le x\le 1$

In order to solve $$e^{x^3-x} =2 $$ We take logarithms to get $$ {x^3-x} = \ln2$$

We are looking for solutions in the closed interval $ [0,1].$

Note that for $ x\in [0, 1]$

$$ {x^3-x} \le 0$$ and $\ln(2) =0.6931...$ is positive.

Thus there is no solution between $0$ and $1$. The option $1$ is the correct answer.

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$e^{x^3-x}=2=e^{ln 2}$

⇒ $x^3-x=ln 2$

$x ≈ 1.245$

Hence this equation has no solution in $[0, 1].$

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A simple way to get the answer is to note that $x^{3}-x\leq 0$ and hence $e^{x^{3}-x} \leq 1 <2$.

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