Let I:= [a, b] and let f: I -> R be a continuous wetion such that f(x) >0 for each x iN I. Prove that here exists a Number of >0 such that f(x) > α for all XEI.

2. Is the proof correct? Either state that it is, or circle the first error and explain what is incorrect about it. If the proof is not correct, can it be fixed to prove the claim true? Claim: If f : (0,1] → R and g : [1,2) → R are uniformly continuous on their domains, and f(1) = g(1), then the function h : (0, 2) → R, defined by h(æ) = { F(x) for x € (0, 1] | 9(x) for x e [1, 2) is uniformly continuous on (0, 2). Proof: Let e > 0. Since f is uniformly continuous on (0, 1], there exists d1 > 0 such that if x, y E (0, 1] and |æ – y| 0 such that if æ, y E [1, 2) and |r – y| < d2, then |9(x) – 9(y)| < €/2. Let 8 = min{d1, &2}. Now suppose r, Y E (0, 2) with x < y and |x – y| < 8. If x, y E (0, 1], then |x – y| < 8 < 81 and so |h(x) – h(y)| = |f (x) – f(y)|< e/2 < e. If x, y € (1,2), then |æ – y| < 8 < d2 and so |h(x) – h(y)| = |9(x) – g(y)| < e/2 < e. If x € (0,1) and y E (1,2), then |x – 1| < |x – y| < 8 < dị and |1 – y| < |æ – y| < 8 < 82 so that |h(x) – h(y)| = |f (x) – f(1) + g(1) –…

2. Let f: I R be a continuous function on a bounded, closed interval I= [a, b). Show that f has a minimum on I.

Use the Mean Value Theorem to prove: If f(x) and g(x) are differentiable on an interval (a, b) and f'(x) = g'(x) for all x in (a, b), then there is a constant k such that g(x) = f(x) + c for all x in (a, b).