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Are (fog)(x) = (gof)(x)? I. Yes. (f °g)(x) = (g°f)(x) = f(x) • g(x) = g • f(x) = (x² – 1) (2x– 4) II. Yes. (fog)(x) = (2x – 4) – 1 = 2x? – 5 and (gof)(x) = 2x² – 1 – 4 = 2x? – 5 III. No. (fog)(x) = (2x – 4)? – 1 = 4x² – 16x +15, but (g•f)(x) = 2(x² – 1) – 4) = 2x² – 6 (13) %3D %3D %3D %3D %3D (A) II only (B) III only (C) I and II (D) None is true

Are (fog)(x) = (gof)(x)? I. Yes. (f °g)(x) = (g°f)(x) = f(x) • g(x) = g • f(x) = (x² – 1) (2x– 4) II. Yes. (fog)(x) = (2x – 4) – 1 = 2x? – 5 and (gof)(x) = 2x² – 1 – 4 = 2x? – 5 III. No. (fog)(x) = (2x – 4)? – 1 = 4x² – 16x +15, but (g•f)(x) = 2(x² – 1) – 4) = 2x² – 6 (13) %3D %3D %3D %3D %3D (A) II only (B) III only (C) I and II (D) None is true

Advanced Engineering Mathematics
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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(13) Are (fog)(x) = (gof)(x)? 1. Yes. (f °g)(x) = (g°f)(x) = f(x) • g(x) = g • f(x) = (x² – 1) (2x, 4) II. Yes. (fog)(x) = (2x – 4) – 1 = 2x? – 5 and (gof)(x) = 2x² – 1 – 4 = 2x² – 5 II. No. (fog)(x) = (2x – 4)? – 1 = 4x² – 16x +15, but (gof)(x) = 2(x² – 1) – 4) = 2x? – 6 (A) II only %3D (B) III only (C) I and II (D) None is true
Transcribed Image Text:(13) Are (fog)(x) = (gof)(x)? 1. Yes. (f °g)(x) = (g°f)(x) = f(x) • g(x) = g • f(x) = (x² – 1) (2x, 4) II. Yes. (fog)(x) = (2x – 4) – 1 = 2x? – 5 and (gof)(x) = 2x² – 1 – 4 = 2x² – 5 II. No. (fog)(x) = (2x – 4)? – 1 = 4x² – 16x +15, but (gof)(x) = 2(x² – 1) – 4) = 2x? – 6 (A) II only %3D (B) III only (C) I and II (D) None is true
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