Derivative Proof

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September 3rd 2009, 08:26 AM
JG89

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
September 3rd 2009, 08:39 AM
swaha

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.
September 3rd 2009, 08:42 AM
HallsofIvy

Quote:

Originally Posted by JG89

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].

Quote:

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
September 3rd 2009, 08:49 AM
JG89

Quote:

Originally Posted by HallsofIvy

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.