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$$\underset{\,\,\,\,{\textcolor[rgb]{0.66,0.66,0.66}{\rightarrow {\mathtt{x}}}}}{{solve}}{\left(\begin{array}{l}{f}{\left({\mathtt{x}}\right)}=\left(\left({\mathtt{x}}{\mathtt{\,\small\textbf+\,}}{\mathtt{2}}\right){\mathtt{\,\times\,}}\left({\mathtt{x}}{\mathtt{\,-\,}}{\mathtt{2}}\right)\right)\\
{f}{\left({\mathtt{x}}\right)}=\left(\left({\mathtt{x}}{\mathtt{\,\small\textbf+\,}}{\mathtt{1}}\right){\mathtt{\,\times\,}}\left({\mathtt{x}}{\mathtt{\,-\,}}{\mathtt{2}}\right)\right)\\
{f}{\left({\mathtt{x}}\right)}=\left(\left({\mathtt{x}}{\mathtt{\,-\,}}{\mathtt{1}}\right){\mathtt{\,\times\,}}\left({\mathtt{x}}{\mathtt{\,-\,}}{\mathtt{1}}\right)\right)\end{array}\right)}$$

 Jan 6, 2015

Best Answer 

 #1
avatar+118608 
+5

It is even if             f( x) = f (-x)

I'll look at the first one.

f(x)=(x+2)(x-2)

 

f(-x)=(-x+2)(-x-2)

f(-x)=-(-x+2)(x+2)

f(-x)=- - (x-2)(x+2)

f(-x)=+ (x+2)(x-2)

f(-x)=f(x)

 

SO YES the first one is even.  :)

 Jan 6, 2015
 #1
avatar+118608 
+5
Best Answer

It is even if             f( x) = f (-x)

I'll look at the first one.

f(x)=(x+2)(x-2)

 

f(-x)=(-x+2)(-x-2)

f(-x)=-(-x+2)(x+2)

f(-x)=- - (x-2)(x+2)

f(-x)=+ (x+2)(x-2)

f(-x)=f(x)

 

SO YES the first one is even.  :)

Melody Jan 6, 2015

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