The equation $e^{x^3-x} - 2 = 0$ has solutions...

Question

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

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)

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. — Arthur, Jan 18 '18 at 06:43
Also "in [0,1]" means "between $0$ and $1$ (including $0$ and $1$)". — Arthur, Jan 18 '18 at 06:45
@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. — dxiv, Jan 18 '18 at 06:49
@DeniKatsman $x^3-x=x(x^2-1) \le 0$ for $x \in [0,1]$, so $e^{x^3-x} \le e^0 = 1 \lt 2$, which would mean that the correct answer is 1.. Maybe you should doublecheck that you asked the right question. — dxiv, Jan 18 '18 at 07:13

score 0 · Accepted Answer · edited Jan 18 '18 at 07:44

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.

score 0 · Answer 2 · answered Jan 18 '18 at 07:48

0

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

⇒ $x^3-x=ln 2$

$x ≈ 1.245$

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

answered Jan 18 '18 at 07:48

sirous

score 0 · Answer 3 · answered Jan 18 '18 at 08:21

0

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

answered Jan 18 '18 at 08:21

Kavi Rama Murthy

3 Answers3