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Math Help - Derivative Proof

  1. #1
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    Derivative Proof

    Could someone please check this proof for me?

    Question: If f(x) is continuous and differentiable in [a,b], show that if  f'(x) \le 0 for  a \le x < \xi and  f'(x) \ge 0 for  \xi < x \le b , the function is never less than  f(\xi) .

    Proof:

    Note that f is continuous in [a,b] and so it must possess a minimum in the interval. Assume that the minimum for the function f in [a,b] occurs at a point x = c such that  c \neq \xi . We can suppose that  a \le c < \xi .

    By hypothesis  f(c) < f(\xi) . And so  f(c) - f(\xi) < 0 . Since  c - \xi < 0 then  \frac{f(c) - f(\xi)}{c - \xi} > 0 . By the MVT there exists a point C in  (c, \xi) such that  f'(C) = \frac{f(c) - f(\xi)}{c - \xi} > 0 , contradicting the fact that f'(x) must be negative or 0 for all x satisfying  a \le x < \xi . The same proof can be applied if we suppose that  \xi < c \le b . QED
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  2. #2
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    yes the proof is correct. but geometrical interpretaions are better & easier for me. the function decreases as the slope is negative from a to epsilon & starts to increase from epsilon to b so at epsilon it is minimum.
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  3. #3
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    Quote Originally Posted by JG89 View Post
    Could someone please check this proof for me?

    Question: If f(x) is continuous and differentiable in [a,b], show that if  f'(x) \le 0 for  a \le x < \xi and  f'(x) \ge 0 for  \xi < x \le b , the function is never less than  f(\xi) .

    Proof:

    Note that f is continuous in [a,b] and so it must possess a minimum in the interval. Assume that the minimum for the function f in [a,b] occurs at a point x = c such that  c \neq \xi . We can suppose that  a \le c < \xi .

    By hypothesis  f(c) < f(\xi) .
    No, that is NOT the hypothesis, that is the conclusion you want to arrive at. What you are doing, I think, is a "proof by contradiction". Suppose the conclusion is NOT true. Then there exist c in [a, b] such that f(c)< f(\xi). Now, apply the mean value theorem to the intervals [a, c] and [c, b].

    And so  f(c) - f(\xi) < 0 . Since  c - \xi < 0 then  \frac{f(c) - f(\xi)}{c - \xi} > 0 . By the MVT there exists a point C in  (c, \xi) such that  f'(C) = \frac{f(c) - f(\xi)}{c - \xi} > 0 , contradicting the fact that f'(x) must be negative or 0 for all x satisfying  a \le x < \xi . The same proof can be applied if we suppose that  \xi < c \le b . QED
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  4. #4
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    Quote Originally Posted by HallsofIvy View Post
    No, that is NOT the hypothesis, that is the conclusion you want to arrive at. What you are doing, I think, is a "proof by contradiction". Suppose the conclusion is NOT true. Then there exist c in [a, b] such that f(c)< f(\xi). Now, apply the mean value theorem to the intervals [a, c] and [c, b].

    I did assume that  f(c) < f(\xi) in the first post.
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