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.
Could someone please check this proof for me?
Question: If f(x) is continuous and differentiable in [a,b], show that if for and for , the function is never less than .
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 . We can suppose that .
By hypothesis . And so . Since then . By the MVT there exists a point C in such that , contradicting the fact that f'(x) must be negative or 0 for all x satisfying . The same proof can be applied if we suppose that . QED
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 . Now, apply the mean value theorem to the intervals [a, c] and [c, b].
And so . Since then . By the MVT there exists a point C in such that , contradicting the fact that f'(x) must be negative or 0 for all x satisfying . The same proof can be applied if we suppose that . QED