Real Analysis Problem?

Question

If f: [a,b] -->R (real number set) is differentiable at c, a < c < b, and f ' (c) > 0, prove that there is x belongs to (c,b) such that
f(x) > f(c).

Answer 1

To refine the first poster's answer: Let ε=f'(c), then by the definition of limit ∃δ>0 s.t. 0<|h|<δ ⇒ |(f(c+h) - f(c))/h - f'(c)| < ε. So choose x = (c + min (b, c+δ))/2. Since both b and c+δ are greater than c, so is their minimum, so c = (c + c)/2 < x < (min (b, c+δ) + min (b, c+δ))/2, min (b, c+δ). Thus in particular x∈(c, b), and also x∈(c, c+δ), so 0<|x-c|<δ. Thus per the previous implication, |(f(x) - f(c))/(x-c) - f'(c)| = |(f(c+(x-c)) - f(c))/(x-c) - f'(c)| < f'(c). Thus (f(x) - f(c))/(x-c) - f'(c) > -f'(c), so (f(x) - f(c))/(x-c) > 0, thus f(x) - f(c) > 0 (since x-c is positive), thus f(x) > f(c). Q.E.D.

Answer 2

If f'(c)>0, then there exists some b>x>c such that
(f(x) - f(c))/(x - c)>k
For some small positive constant k. (This result is straight from definition.)
Now multiply both sides by (x - c)
f(x) - f(c)>k(x - c)
f(x)>k(x - c)+f(c)

Answer 3

If f'(c)>0, then there exists some b>x>c such that (f(x) - f(c))/(x - c)>ok For some small smart consistent ok. (This result's straight removed from definition.) Now multiply the two area by (x - c) f(x) - f(c)>ok(x - c) f(x)>ok(x - c)+f(c)