Advanced Engineering Mathematics
10th Edition
ISBN: 9780470458365
Author: Erwin Kreyszig
Publisher: Wiley, John & Sons, Incorporated
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- Class: Mathematical Analysis/Real Analysis; C2arrow_forwardAdvanced Calculus: Suppose f and g are differentiable functions on an interval I. Show that iff′(x) = g′(x) for all x ∈ I and f(x0) = g(x0) for some x0 ∈ I, then f(x) = g(x) forall x ∈ I.arrow_forward2. 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) –…arrow_forward
- Class: Mathematical Analysis/Real Analysis; C1Aarrow_forwardContinuity at an open interval implies integrability and differentiability at the same open interval. O True O False Suppose that f is differentiable on some open interval containing c, and (c,f(c)) is a point of inflection of the graph of f. Then f"(c)=O. O True O Falsearrow_forwardMATHMATICSarrow_forward
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