1. ## Derivative Proof

Could someone please check this proof for me?

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

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 $\displaystyle c \neq \xi$. We can suppose that $\displaystyle a \le c < \xi$.

By hypothesis $\displaystyle f(c) < f(\xi)$. And so $\displaystyle f(c) - f(\xi) < 0$. Since $\displaystyle c - \xi < 0$ then $\displaystyle \frac{f(c) - f(\xi)}{c - \xi} > 0$. By the MVT there exists a point C in $\displaystyle (c, \xi)$ such that $\displaystyle f'(C) = \frac{f(c) - f(\xi)}{c - \xi} > 0$, contradicting the fact that f'(x) must be negative or 0 for all x satisfying $\displaystyle a \le x < \xi$. The same proof can be applied if we suppose that $\displaystyle \xi < c \le b$. QED

2. 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.

3. 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 $\displaystyle f'(x) \le 0$ for $\displaystyle a \le x < \xi$ and $\displaystyle f'(x) \ge 0$ for $\displaystyle \xi < x \le b$, the function is never less than $\displaystyle f(\xi)$.

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 $\displaystyle c \neq \xi$. We can suppose that $\displaystyle a \le c < \xi$.

By hypothesis $\displaystyle f(c) < f(\xi)$.
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 $\displaystyle f(c)< f(\xi)$. Now, apply the mean value theorem to the intervals [a, c] and [c, b].

And so $\displaystyle f(c) - f(\xi) < 0$. Since $\displaystyle c - \xi < 0$ then $\displaystyle \frac{f(c) - f(\xi)}{c - \xi} > 0$. By the MVT there exists a point C in $\displaystyle (c, \xi)$ such that $\displaystyle f'(C) = \frac{f(c) - f(\xi)}{c - \xi} > 0$, contradicting the fact that f'(x) must be negative or 0 for all x satisfying $\displaystyle a \le x < \xi$. The same proof can be applied if we suppose that $\displaystyle \xi < c \le b$. QED

4. 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 $\displaystyle f(c)< f(\xi)$. Now, apply the mean value theorem to the intervals [a, c] and [c, b].

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