# mean value theorem or intermediate value theorem or extreme value theorem?

I submitted this assignment and the teacher gave me 1/5 marks for this and I dont know why, she just said not enough information to prove the point. She NEVER tells you the solution even after she has marked the assignments so I feel pretty miserable because I dont know why my answer is not good enough. So I am asking here in hopes someone can help explain this to me.

The question is: A hiker leaves his cabin at 8 am and walks uphill to another cabin and reaches that cabin uphill at 8 pm. the next morning he leaves from his uphill cabin to go back to his downhill cabin. he leaves at 8 am and reaches the downhill cabin at 8 pm. prove that at one point in time the hiker was at the exact same spot when going uphill as he was downhill.

Now what I wrote for my solution was basically a graph of the hiker's journey, the graph looks kind of like $-x^2$'s graph. I said in order for him to make it to the cabin uphill or down hill, he must have maintained the same AVERAGE velocity. because the average velocity was the same for both trips, he must have reached one point at the exact same time on both trips.

she said not enough info and gave me 1/5..didnt even look at my beautiful graph or explanation

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I admit, I find it an incomplete explanation as well. –  mixedmath Dec 9 '11 at 17:21
@Nadal: Even if your teacher did not write a complete explanation on your homework, you can still ask her what went wrong. Writing detailed solutions on each assignment is impractical, especially if you have a large number of students. –  JavaMan Dec 9 '11 at 17:30
she doesnt post solutions either...now what.. –  Raynos Dec 9 '11 at 20:06

I agree with her assessment: your answer doesn’t get at the real reason, and the actual graph needn’t look at all like that of $f(x)=-x^2$: the hiker might have stopped along the way, or even turned and gone partway down the hill a few times.

The real key is the intermediate value theorem. Let $f(t)$ be his distance from the lower cabin at time $t$ on the first day, and let $g(t)$ be his distance from the lower cabin at time $t$ on the second day. Graph $y=f(t)$ and $y=g(t)$ on the same plot. If $d$ is the distance between the two cabins, and $t$ is measured in hours, $f(0)=g(12)=0$ and $f(12)=g(0)=d$. It’s intuitively clear that these two graphs must cross somewhere, because $f$ and $g$ are continuous. To make that precise, look at the function $h(t)=g(t)-f(t)$: it’s also continuous, $h(0)=g(0)-f(0)=d$, and $h(12)=g(12)-f(12)=-d$, so the intermediate value theorem tells you that there must be some $t_0$ between $0$ and $12$ such that $h(t_0)=0$. But then $g(t_0)-f(t_0)=0$, and hence $g(t_0)=f(t_0)$: at time $t_0$ on each day he was distance $f(t_0)$ from the lower cabin.

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If he goes the same distance in the same time, then he must have had the same average velocity, so that part of what you wrote is correct.

In one sense the simpler way to think about this scenario is to think about one person going upward and the other going downward at the same time. They pass each other. At that point they're at the same place at the same time.

That's the intermediate value theorem. If the height of one of the two walkers above the other walker goes from positive to negative, then at some point in between it's zero.

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Suppose $f(t)$ is how far from the cabin he is at time $t$ on the first day, and $g(t)$ is how far from the cabin he is at the time $t$ on the second day.

Then note that both $f$ and $g$ are both continuous, so $f - g$ is continuous. Also, $f(8\text{ am}) - g(8\text{ am}) < 0$, and $f(8\text{ pm}) - g(8\text{ pm}) > 0$. By the intermediate value theorem, there is at least one time when the function takes the value 0. This means that there is a time when $g(t) = f(t)$.

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