Rolle's theorem and concav/convex function

Question

Let’s say we have $g(x)=(1+x)e^{-x}$

1) Can we say that the function is concav ? convex ? neither convex nor concav ?

This is the wording of the exam.

$g''(x)=(x-1)e^{-x}$

So, normally, we can tell that $g(x)$ is concav between $\left[-\infty;1 \right]$ and convex elsethere.

The fact that it is not concav everywhere, does it mean that we can tell that it is neither convex nor concav ?

2) $g(x)=0$ has a unique solution. True or false ?

We know that there exists at least one solution, which is -1. We know that this should be the only solution as the exponentiel does not vanish.

By Rolle’s Theorem, there exists more than one solution if $g’(x)$ has a root. Since $g'(x)=-xe^{-x}$ has a root, which is $0$, we can say there are more than one solution.

Can somebody help to use Rolle's theorem to prove that there is at most 1 solution to $g(x)=0$ ?

Addition: Can we use the Rolle's theorem here ? I am not so sure, given that the question does not refer to a closed interval ?

Edit: I rephrased the question of 2)

Rolle's theorem guarantees a zero of $g'$, given two zeros of $g$, whereas you are trying to stipulate the converse, which is wrong. — Christian Blatter, Aug 08 '14 at 08:49

score 3 · Accepted Answer · answered Aug 07 '14 at 15:28

3

1) is correct.

For 2) use the fact $ab=0\iff (a=0)\lor( b=0)$ and that the exponential function doesn't vanish.

answered Aug 07 '14 at 15:28

Sorry, but that does not answer the question of uniqueness of solution that I am referring to. I know there is one solution, but I want to know if there is another solution(s). – XCoder Aug 07 '14 at 20:22
1

Hi Sami. Have a great hours ahead. :) – Mikasa Aug 08 '14 at 14:02

Rolle's theorem and concav/convex function

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