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- May 7th 2007, 06:17 PM #1

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- May 7th 2007, 06:21 PM #2

- May 7th 2007, 06:27 PM #3

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If y'=0 on I some open interval.

Then, y=C for some real number C.

**Proof:**Let a<b be in I. Then consider [a,b]. The function y is differenciable fully on [a,b] because it is differenciable on I. Thus, y is continous on [a,b] and as said on (a,b). This tells us that y satisfies Lagrange's Mean Value Theorem. Hence there exists c in (a,b) so that,

f'(c)=(f(b)-f(a))/(b-a)

But f'(c)=0 because f'(x)=0 for all x in I.

Thus,

0=(f(b)-f(a))/(b-a)

Thus,

f(b)=f(a).

This tells us that f is a konstant function on I.

Q.E.D.

- May 7th 2007, 06:28 PM #4

- May 7th 2007, 06:31 PM #5

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- May 7th 2007, 06:34 PM #6

- May 7th 2007, 07:01 PM #7

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i should i stated the whole problem. im sorry. it went as follows:

For what value(s) of x in this equation does y attain its minimum value if a, b, and c are all positive constants?

y=[(a^2)+(b-x)^2]^(1/2) + [(c^2)+(x^2)]^(1/2)

and someone else told me:

Put a=1, b=2, c=3. Then plot the curve. Is the minimum at x=2 or x=0?

-its not.

so, This curve is smooth and y can be minimised by solving dy/dx=0 in the usual manner. I just dont know what the "usual manner" is..