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Using the IVT and Rolle's Theorem, prove that the equation $$e^x-x-2=0$$ has exactly one positive real solution. Use the IVT to prove that the equation has at least one positive real solution. Using proof by contradiction and Rolle's Theorem show that the equation has exactly one positiove real solution.

This is what I have so far. Please correct me if i have made a mistake.

Define function $f:[1,6]\rightarrow\mathbb{R}$ by $f(x)=e^x-x-2$. Function $f$ is continuous since $e^x$ and $-x-2$ are continuous functions so by the algebra of continuous functions $f(x)$ is a continuous function.

Notice that $f(1)<0$ and $f(6)>0$. So $f(6)<0<f(1)$. Applying the IVT, we have that there exists $c\in(1,6)$ such that $f(c)=0$ $\iff$ $e^c-c-2=0$. Thus the equation $e^x-x-2=0$ has at least one positive (since our interval is positive only) real solution. How do I prove that there is exactly one by Rolle's?

snowman
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Assume that there are two positive solutions $a,b.$ Then $f:[a,b]\to \mathbb{R}$ is continuous on $[a,b],$ derivable on $(a,b)$ and $f(a)=f(b)=0.$ Then, by Rolle's theorem there exists $c\in(a,b)$ such that $f'(c)=0.$ But, since $c>a>0,$ it is

$$f'(c)=e^c-1>e^0-1=0, \forall c\in (0,\infty),$$ what is a contradiction. So, we can conclude that there are not two positive solutions.

mfl
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  • no im sure it is $e^x-x-2$. It did say positive solutions though... – snowman Dec 04 '14 at 16:41
  • how do you know that your f is continuous on [a,b] and diff on (a,b)? – snowman Dec 04 '14 at 20:59
  • $f(x)=e^x-x-2$ is continuous, as you have said, and differentiable with $f'(x)=e^x-1.$ – mfl Dec 04 '14 at 21:05
  • how come you never said what you function is defined by? like the rule that assigns elements from domain to codomain. like in mine it was the f(x). – snowman Dec 04 '14 at 22:28
  • I didn't mention it because I was using the same function as in your question. – mfl Dec 04 '14 at 22:31
  • so you were using the same function but just different interval right? – snowman Dec 04 '14 at 22:32
  • Yes. The interval given by two possible roots. The proof of this kind of problems using Rolle is by contrapositive. You assume there is two roots. Then, by Rolle, there is a point where the derivative vanishes. If the derivative never vanishes, this makes no sense. Since the only assumption is the existence of two roots (positive in this particular case) the nonsense of the conclusion implies that the assumption doesn't hold. That is, $f$ has no two positive roots. – mfl Dec 04 '14 at 22:35
  • also, this might be a dumb question but you concluded that there is not two positive solutions. but that doesn't rule out the possibility of there being three or four or five... solutions does it? – snowman Dec 04 '14 at 23:28
  • The argument works if you assume the existence of, at least, two roots. So this argument excludes the possibility of a higher number of roots. – mfl Dec 04 '14 at 23:50
  • in your last equation, shouldn't you also say that this is true since e^x-1 is a strictly increasing function $\forall x\in\mathbb{R}$? – snowman Dec 06 '14 at 13:51