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(Click the picture to see the problem and what I have done so far)

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- Sep 24th 2012, 03:12 PMChaimUsing IVT (Int Value Therom) on a Hill
Attachment 24920

(Click the picture to see the problem and what I have done so far) - Sep 24th 2012, 04:15 PMTheEmptySetRe: Using IVT (Int Value Therom) on a Hill
So you have defined two different functions. You have

$\displaystyle s:[0,20] \to [0,1]$

and

$\displaystyle r:[0,10] \to [0,1]$

For these functions to be compatable they must agree on where you define the top and the bottom.

If you defince the top to 1 and the bottom to be 0 then you must have these values

$\displaystyle s(0)=0 \quad s(20)=1$

and

$\displaystyle r(0)=1 \quad r(10)=0$

So you get the two functions

$\displaystyle s(t)=\frac{t}{20} \quad r(t)=1-\frac{t}{10}$

Now if you define their difference it will only exist on the intersection of their domains!

$\displaystyle f:[0,10] \to [-1,.5] \quad f(t) =s(t)-r(t)=\frac{3t}{20}-1$

But notice that

$\displaystyle f(0) =-1 \text{ and } f(10)=.5$

Since f is continous the intermediate value theorem says there exists a $\displaystyle c \in [0,10]$ such that $\displaystyle f(c)=0$

So we get that

$\displaystyle f(c)=s(c)-r(c) =0 \iff 0=s(c)-r(c) \iff s(c)=r(c)$

So the hiker was at the same place at the same time.

Note: Since the function is linear would could solve for the value of t explicitly, but the power of the IVT is we don't have to be able to find it.