How do you determine if #f(x)=sinxsqrt(x²+1)# is an even or odd function?

1 Answer
Parabola · Rishi A.
Mar 19, 2016

#f(x)# is an odd function.

Explanation:

If #f(-x)=f(x)#, it is an even function.
If #f(-x)=-f(x)#, it is an odd function.
If neither equation is true, the function is neither even nor odd.

#f(-x)#
#=sin(-x)sqrt((-x)^2+1)#
#=-sin(x)sqrt((-x)^2+1)#
#=-sin(x)sqrt(x^2+1)#
#=-f(x)#

Thus, #f(x)# is an odd function.

Now, what does it mean by a function being even or odd?

Well, if a function is even, it is symmetric to the #y# axis.
Well, if a function is odd, it is symmetric to the origin.

What if a function were to be symmetric to the #x# axis?
An example for this would be #y=+-sqrtx#. This is indeed symmetric to the #x# axis, but it has two #y# values for a given #x# value. This means that it is not a function.

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