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# is 2^x a even function

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is 2^x a even function

Jul 29, 2014

#1
+4472
+13

if f(x) = f(-x), then the function is even. So, f(-x) = 2^-x which does not equal 2^x.

However, if f(-x) = -f(x), then the function is odd. So -f(x) = -2^x which does not equal f(-x) = 2^-x.

f(x) = 2^x is neither even nor odd.

Jul 29, 2014

#1
+4472
+13

if f(x) = f(-x), then the function is even. So, f(-x) = 2^-x which does not equal 2^x.

However, if f(-x) = -f(x), then the function is odd. So -f(x) = -2^x which does not equal f(-x) = 2^-x.

f(x) = 2^x is neither even nor odd.

AzizHusain Jul 29, 2014
#2
+95177
+10

Also,

A function that is even has the y axis (x=0) as an axis of symmetry       f(x)=f(-x)

A function that is odd has point symmetry about the origin. This means that if it is rotated 180 degrees it will be the same.     f(-x)=-f(x)

This is what 2x looks like.  You can see that it is not symmetrical about the y axis.

And it does not have point symmetry about (0,0)

so it is not an even function or an odd function.

Jul 30, 2014
#3
+94526
+5

Let me add one thing to what Melody is saying......we can always "test" whether a function is even by replacing x with -x.....if the reults are the same, then the function is even.

Then, is 2(x) = 2(-x)  ???........NO !!!

Note that a function like x2 is even, because...

x2 = (-x)2

And this is exactly what Melody said......

Jul 30, 2014
#4
+95177
0

Thanks Chris,

Jul 30, 2014
#5
+94526
0

I saw that after I had posted...let me give Aziz full credit too!!   (I won't take any points for my repeated answer!!.......)

Jul 30, 2014
#6
+95177
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You do not need to take points off yourself Chris, I thought your answer was worth a tick!

It was a good point to reiterate - and yours was perhaps a little clearer.

Jul 30, 2014