# Derivative Proof

• Sep 3rd 2009, 09:26 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
• Sep 3rd 2009, 09:39 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.
• Sep 3rd 2009, 09:42 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
• Sep 3rd 2009, 09:49 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.