# Using IVT and Rolle's Theorem

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?

• Suppose there were two positive solutions $c_1,c_2$. What does Rolle's theorem say about that? – Daniel Fischer Dec 4 '14 at 16:20
• Are you sure that it is $f(x)=e^x-x-2?$ It has two different roots. Maybe, it is $e^x+x-2.$ – mfl Dec 4 '14 at 16:32
• @mfl $f(x) = e^x - x - 2$ has only one positive real zero. – Daniel Fischer Dec 4 '14 at 16:41
• @DanielFischer I have not read positive. Thank you for noticing. – mfl Dec 4 '14 at 17:53

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.
• no im sure it is $e^x-x-2$. It did say positive solutions though... – snowman Dec 4 '14 at 16:41
• $f(x)=e^x-x-2$ is continuous, as you have said, and differentiable with $f'(x)=e^x-1.$ – mfl Dec 4 '14 at 21:05