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Let $f(x)=e^{-x}$ and $g(x)=\frac{1}{\sqrt{2\pi}}e^{-\frac{(x-1)^2}{2}}+\frac{1}{\sqrt{2\pi}}e^{-\frac{(x+1)^2}{2}}$. To complete an exercise I had to show there exists $x_1<x_2$ such that $g(x)\le f(x)$ for $x<x_1$, $g(x)\ge f(x)$ for $x_1<x<x_2$ and $g(x)\le f(x)$ for $x>x_2$. From enter image description here this picture it seems to be true. If I try to plot $h(x):=f(x)-g(x)$ I get enter image description here which has $2$ roots. Is there anyway to show this without CAS?

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    $\begingroup$ Use the Intermediate Value Theorem and some clever guesses. $\endgroup$ – Sean Roberson Feb 23 at 7:58
  • $\begingroup$ To answer this problem you don't need to find the points where the curves cross (in the sense of determining their values exactly). You just need to show that they exist (and no others). The standard tools are the I.V.P. for existence and the M.V.P. for uniqueness. $\endgroup$ – Sammy Black Feb 23 at 8:02
  • $\begingroup$ I'm quite new to those theorems. Is there any way you could give some hints how to approach it? $\endgroup$ – statman Feb 23 at 8:24
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    $\begingroup$ @SammyBlack So since $h(-1)=2.27, h(1)=-0.09$ and $h(4)=0.01$ by IVP we know that there is a root in $[-1,1]$ and a root in $[1,4]$. How to I show that they are the only ones? $\endgroup$ – statman Feb 23 at 9:01
  • $\begingroup$ I just realized that abbreviated Intermediate Value Theorem as I.V.P. because I've been teaching differential equations and using those initials so often for Initial Value Problem. Then, by muscle memory, I also abbreviated Mean Value Theorem as M.V.P. Whoops! Those should be I.V.T. and M.V.T. Sorry. $\endgroup$ – Sammy Black Feb 23 at 9:41
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This is a response to the question raised in the comments: how to use the I.V.T. and M.V.T. It's not a complete answer.

Using the Intermediate Value Theorem to prove existence of a root of a continuous function $h$ within an interval $[a, b]$: check that $h(a)$ and $h(b)$ have opposite signs. Since $0$ lies between those values, the theorem guarantees the existence of a number $c \in (a, b)$ such that $h(c) = 0$.

Using the Mean Value Theorem to prove uniqueness of a root $c$ of a differentiable function $h$ within an interval $[a, b]$: use a contradiction argument. Call this root $c_1$ and assume that there's a second root $c_2$ within the interval. Then the theorem guarantees the existence of a number $d \in (c_1, c_2)$ where the tangent line is parallel to the secant line through those two roots, i.e. $$ h'(d) = \frac{h(c_2) - h(c_1)}{c_2 - c_1} = \frac{0 - 0}{c_2 - c_1} = 0. $$ Then you calculate the derivative and use some inequality to show that it does not take on the value $0$ on the interval $(a, b)$ which contains $(c_1, c_2)$. This technique allows you to verify that the derivative is nonzero in place of verifying that the original function is nonzero. Of course, there's no guarantee that this is easier!

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  • $\begingroup$ Ah okay that makes sense. But if I know there is only one root in $[-1,1]$ and and only one root in $[1,4]$ how do I check if there isn't a root in $(4,\infty)$? For $x<-1$ it is each it is easy to check that $h$ will also be positive but for $x>4$ it hovers around $0$. $\endgroup$ – statman Feb 23 at 10:13
  • $\begingroup$ You're probably going to have to make a direct asymptotic argument about how the exponentials with quadratic arguments decay faster than the exponential with the linear argument. $\endgroup$ – Sammy Black Feb 23 at 10:53

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