BIG IDEAS MATH Algebra 1: Common Core Student Edition 2015

1st Edition

ISBN: 9781608408382

Author: HOUGHTON MIFFLIN HARCOURT

Publisher: Houghton Mifflin Harcourt

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Question

Chapter 8.4, Problem 76E

(a)

To determine

A constant multiple of an even function being even.

(a)

Expert Solution

Answer to Problem 76E

$g(−x)=g(x)$

Explanation of Solution

Given information:

Any constant multiple of an even function is even.

Formula used:

$g(−x)=af(−x)$

Calculation:

True

Let $f(x)$ be an even function. Let $g(x)=af(x)$ where $a∈ℝ$

$g(−x)=af(−x)$

Since f is even function, we get

$g(−x)=af(x) =g(x)$

Thus, g is an even function.

Conclusion:

$g(−x)=g(x)$

(b)

To determine

A constant multiple of an odd function being odd.

(b)

Expert Solution

Answer to Problem 76E

$g(−x)=−g(x)$

Explanation of Solution

Given information:

Any constant multiple of an odd function being odd.

Formula used:

$g(−x)=af(−x)$

Calculation:

True

Let $f(x)$ be an even function. Let $g(x)=af(x)$ where $a∈ℝ$

$g(−x)=af(−x)$

Since f is even function, we get

$g(−x)=af(−x) =−g(x)$

Thus, g is an even function.

Conclusion:

$g(−x)=−g(x)$

(c)

To determine

The sum or difference of two even function is even.

(c)

Expert Solution

Answer to Problem 76E

$g(−x)=f(x)+ah(x) =g(x)$

Explanation of Solution

Given information:

The sum or difference of two even function is even.

Formula used:

$g(−x)=f(−x)+ah(−x)$

Calculation:

True

Let $f(x)$ be an even function. Let $g(x)=f(x)+ah(x)$ where $a∈{−1,+1}$

$g(−x)=f(−x)+ah(−x)$

Since f, h is even function, we get

$g(−x)=f(x)+ah(x) =g(x)$

Thus, g is an even function.

Conclusion:

$g(−x)=f(x)+ah(x) =g(x)$

(d)

To determine

The sum or difference of two odd functions is odd.

(d)

Expert Solution

Answer to Problem 76E

$g(−x)=−f(x)+a(−h(x)) =−g(x)$

Explanation of Solution

Given information:

The sum or difference of two odd functions is odd.

Formula used:

$g(−x)=f(−x)+ah(−x)$

Calculation:

True

Let $f,h$ be an even function. Let $g(x)=f(x)+h(x),a∈{−1,+1}$

$g(−x)=f(−x)+ah(−x)$

Since f, h are odd function, we get

$g(−x)=−f(x)+a(−h(x)) =−g(x)$

Thus, g is an odd function.

Conclusion:

$g(−x)=−f(x)+a(−h(x)) =−g(x)$

(e)

To determine

The sum or difference of an even function and odd function is odd.

(e)

Expert Solution

Answer to Problem 76E

g is not odd function.

Explanation of Solution

Given information:

The sum or difference of an even function and an odd function is odd.

Formula used:

$f(x)=x2 −f is even functionh(x)=x2 −h is odd functiong(x)=f(x)+h(x)$

Calculation:

False

A counter-example is following:

$f(x)=x2 −f is even functionh(x)=x2 −h is odd functiong(x)=f(x)+h(x)$

Note that

$f(1)=h(1)=1, g(1)=2≠−g(−1)=0$

Thus, g is not odd function.

Conclusion:

g is not odd function.

Chapter 8.4, Problem 75E

Chapter 8.4, Problem 77E

Chapter 8 Solutions

BIG IDEAS MATH Algebra 1: Common Core Student Edition 2015

Ch. 8.1 - Prob. 1E Ch. 8.1 - Prob. 2E Ch. 8.1 - Prob. 3E Ch. 8.1 - Prob. 4E Ch. 8.1 - Prob. 5E Ch. 8.1 - Prob. 6E Ch. 8.1 - Prob. 7E Ch. 8.1 - Prob. 8E Ch. 8.1 - Prob. 9E Ch. 8.1 - Prob. 10E

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