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Calculus, 8th Edition,stewart; page 23 problem 79

If f and g are both even functions, is f + g even ? if f and g are both odd functions, is f + g odd ? What if f is even and g is odd ? justify your answers.

asked Jul 28, 2015 in CALCULUS by anonymous

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1 Answer

Step:1

Consider $f$ and $g$ are the even functions.

From the definitions $f(-x)=f(x)$ and $g(-x)=g(x)$ .

Find the addition $f+g$ .

$(f+g)(-x)=f(-x)+g(-x)$

$(f+g)(-x)=f(x)+g(x)$

$(f+g)(-x)=(f+g)(x)$ .

If $f$ and $g$ are the even functions, then the addition $f+g$ is an even function.

Step:2

Consider $f$ and $g$ are the odd functions.

From the definitions $f(-x)=-f(x)$ and $g(-x)=-g(x)$ .

Find the addition $f+g$ .

$(f+g)(-x)=f(-x)+g(-x)$

$(f+g)(-x)=\left [ (-f(x))+(-g(x)) \right ]$

$(f+g)(-x)=-\left [f(x)+g(x) \right ]$

$(f+g)(-x)=-(f+g)(x)$ .

If $f$ and $g$ are the odd functions, then the addition $f+g$ is an odd function.

answered Jul 28, 2015 by dozey Mentor

Step:3

Consider $f$ is an even function and $g$ is an odd function.

From the definitions $f(-x)=f(x)$ and $g(-x)=g(x)$ .

Find the addition $f+g$ .

$(f+g)(-x)=f(-x)+g(-x)$

$(f+g)(-x)=f(x)+(-g(x))$

$(f+g)(-x)=f(x)-g(x)$

This is not equal to either $(f+g)(x)$ or $-(f+g)(x)$ .

Thus, $f+g$ is neither even nor odd function.

Solution:

If $f$ and $g$ are the even functions, then the addition $f+g$ is an even function.

If $f$ and $g$ are the odd functions, then the addition $f+g$ is an odd function.

If one of the function is even and other is odd, then the addition $f+g$ is neither even nor odd function.

commented Jul 28, 2015 by dozey Mentor

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