StatementRelated factsFacts usedProof
Step no. |
Assertion/construction |
Facts used |
Given data used |
Previous steps used |
Explanation
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1 |
If is zero on all of , then for any choice of |
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obvious
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2 |
must attain both its maximum and its minimum values on |
Fact (1) |
is continuous on |
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Given-fact combination direct
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3 |
If is not zero on all of , either its absolute maximum value on is positive and attained at a point in the open interval or its absolute minimum value on is negative and attained at a point in the open interval (or possibly both). |
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Step (2) |
[SHOW MORE]If  is not everywhere zero, it must either attain positive values or negative values somewhere in  . By Step (2), the function attains both its maximum and minimum values. If the function attains positive values, then the maximum value is positive, hence must be attained at some point in the interior. If the function attains negative values, then the minimum value is negative, hence must be attained at some point in the interior.
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4 |
If is a point in at which attains its maximum value or its minimum value, then . |
Fact (2) |
is differentiable on |
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[SHOW MORE]A maximum value (respectively minimum value) in the interior is also a local maximum value (respectively, local minimum value) for the function, so by Fact (2),  is a critical point for  . Thus, either  or  does not exist. The does not exist case cannot arise because  is given to be differentiable on  . Thus, we are forced to have  .
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5 |
There is a point at which . |
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Steps (1), (3), (4) |
[SHOW MORE]Step (1) settles the case of the zero function. If  is not the zero function, Step (3) says that  attains either its maximum value or its minimum value at some interior point. Step (4) now tells us that the derivative at that point is zero, completing the proof.
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