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You've got a job at a company and you will be traveling to a conference with accommodation. You leave home on monday at 07:15 and arrive at 11:07. The next day you drive exactly the same route back. There is less traffic so you start at 8:32 and arrive at 11:01. Show that there is a point on that route you were at, at the exact same time the two days.

Okay, so that is my question. Does anyone know how to go about this one? I think I need to use the intermediate value theorem, but I'm not sure how to! Thanks in advance for tips/solutions.

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  • $\begingroup$ Intuitively, you could imagine that there are two divers: one traveling to the conference and one traveling back. You can see that It does not matter how fast they are traveling and that they will run into each other eventually. $\endgroup$
    – recmath
    Sep 2, 2014 at 0:20

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Extend your drive back: get in the care at 7:15 and wait until 8:32 to start moving; when you arrive at 11:01, sit in the car until 11:07.

The fraction of the way you've gone along the route on the first day is a function $$ f:[7:15, 11:07] \to [0, 1] $$ with $f(7:15) = 0$ and $f(11:07) = 1$.

Similarly, there's a function $g$ indicating the fraction for the return trip, but for $g$, we have $g(7:15) = 1$ and $g(11:07) = 0$.

Let $h = g - f$. Then $h$ is continuous, $h(7:15) > 0$, and $h(11:07) < 0$. Now you should be able to apply the IVT.

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  • $\begingroup$ hmm, and how do I do that? $\endgroup$
    – 123sfee987
    Sep 1, 2014 at 23:56
  • $\begingroup$ Well, $h$ is a continuous function on a closed interval in the real numbers (once you convert "time of day" to something like "minutes past midnight"); it's positive at one end of the interval, and it's negative at the other. What does the IVT tell you about $h$? $\endgroup$
    – John Hughes
    Sep 1, 2014 at 23:57
  • $\begingroup$ In fact, let me ask another question, just to be sure we're talking about the same thing: what's the statement of the Intermediate Value Theorem? $\endgroup$
    – John Hughes
    Sep 2, 2014 at 0:01
  • $\begingroup$ if s is a number between f(a) and f(b) then there exist a c in [a,b] such that f(c)=s? $\endgroup$
    – 123sfee987
    Sep 2, 2014 at 0:13
  • $\begingroup$ Close. You need to know that the entire interval $[a, b]$ is in the domain of $f$, and that $f$ is continuous on that domain. Now apply that theorem to my function $h$, with $s = 0$ (which is between the two values at the endpoints, which are $-1$ and $1$. What's it tell you? $\endgroup$
    – John Hughes
    Sep 2, 2014 at 3:34

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