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Explain why or why not Determine whether the | Ch 1.1 - 55E

Calculus: Early Transcendentals | 1st Edition | ISBN: 9780321570567 | Authors: William L. Briggs, Lyle Cochran, Bernard Gillett ISBN: 9780321570567 2

Solution for problem 55E Chapter 1.1

Calculus: Early Transcendentals | 1st Edition

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Calculus: Early Transcendentals | 1st Edition | ISBN: 9780321570567 | Authors: William L. Briggs, Lyle Cochran, Bernard Gillett

Calculus: Early Transcendentals | 1st Edition

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Problem 55E

Explain why or why not Determine whether the following statements are true and give an explanation or counterexample. a. The ?ra?nge? of? ?f? ) ? 2x? ? 38 is all real numbers. b. ? h?e rel?ation f? ? ) ? ? 6+ 1 is n? ot a funct?ion bec ? ause f? (1) = ? (? 1) = 2. ? ? ? c. If?? ? )? ? ?1,? t? en ?f(1/?x) = 1?/f? ). ? ? ? d. In? g?eneral, f ? (?f(?x)) =? ? ))2. ? ? ? e.? I?n ?general, ? (?? )) ?= g(f? ? )). ? ? ? ? f.? In?general, f ? (?g? )) = (?f ° ?g)? ). ? ? g. If ? ?? )is? an even function, then ?c f(?ax?)is an?even function, where a ? and ?c are real numbers. ? ? ?h.? If ?? )is an odd function, then f ? (? )? d is an odd function, where d ? is a real number. ? ? ? ? i.?If ?f is both even a ? nd odd, then ?f(x ? ) = 0 for all x ? .

Step-by-Step Solution:

Step-by-step solution Step 1 of 9 (a) Different functions have defined domain by its definition. In this case f is polynomial function and the range of all polynomial functions is all reals numbers . Therefore, the answer is true. Step 2 of 9 (b) By the definition of function every function can have the same value for different values of x, but it cannot have different values for one value of x. In this case we have two values of x, 1 and -1, and one value of f(x), 2. Therefore f is function and the statement is not true. For this function we have the following: 6 f(x) = x +1 f(x) = (x) +1 = f(x) Therefore the function is even. Based on all of this we can say that the answer is false. Step 3 of 9 (c) The function is (x) = x-1. Let’s find (1/x): f(x) = x1 f( ) = 1 x x1 x = x x = 1x x = xx1 f(x) = x Therefore, the answer is false. Step 4 of 9 (d) Let’s apply simple analysis: If f(x)=x then: f(f(x)) = f(x) = x 2 Therefore f(f(x)) =/ (f(x)) . Based on this we can say that the statement is false. Step 5 of 9 (e) Let’s apply simple analysis: If f(x) = x 2 and g(x) = 1+x then we have: 2 f(g(x)) = (g(x)) 2 2 = (1+x) 1 = 1 +2·1·x+x 2 (Since (a+b) = a +2ab+b ) 2 2 = 1+2x+x 2 2 2 = x +2x1 = x(2+x) Let’s now find g (f(x)): g(f(x)) = 1+f(x) = 1+x 2 = x 1 Based on the previous we can conclude that f(g(x)) = / g(f(x)). Therefore the answer is false.

Step 6 of 9

Chapter 1.1, Problem 55E is Solved
Step 7 of 9

Textbook: Calculus: Early Transcendentals
Edition: 1
Author: William L. Briggs, Lyle Cochran, Bernard Gillett
ISBN: 9780321570567

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