Please provide formal proof for both 1.1 and 1.21.1Let f and g be real-valued C functions on [a, b]. Assume f

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Let f and g be real-valued C' functions on [a, b]. Assume f < g on [a, b]. Define S = {(x, y) E R² : a < x< b, f (x)< y

Advanced Engineering Mathematics

10th Edition

ISBN:9780470458365

Author:Erwin Kreyszig

Publisher:Erwin Kreyszig

Chapter2: Second-order Linear Odes

Section: Chapter Questions

Problem 1RQ

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This material is Real analysis.
Determine whether the following statements are true or false. If true, provide a proof; if false, provide a counterexample. (a) Let f, g, h be continuous on the interval [a, b]. If f(a) g(b) > h(b), then there exists c = [a, b] such that f(c) = g(c) = h(c). (b) Suppose that f and g are continuous on R. If 0 ≤ f(x) < g(x) for all X, then there is some x ER such that f(x)/g(x) is the maximum value of f/g. (c) If f is continuous on R, then f is bounded.
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) –…
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