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

Question

Ariana Alvarez · Accepted Answer

#f(x)# is an odd function.

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.

Ariana Alvarez · Answer

undefined To determine if $ f(x) = \sin(x)\sqrt{x^2 + 1} $ is an even or odd function, we need to examine its symmetry properties with respect to the y-axis (even function) or origin (odd function).

1. Even function: $ f(x) $ is even if $ f(-x) = f(x) $ for all $ x $ in the domain of $ f(x) $.
2. Odd function: $ f(x) $ is odd if $ f(-x) = -f(x) $ for all $ x $ in the domain of $ f(x) $.

Let's check:

$$
f(-x) = \sin(-x)\sqrt{(-x)^2 + 1} = -\sin(x)\sqrt{x^2 + 1}
$$

Comparing this to $ f(x) $, we see that $ f(-x) = -f(x) $. Therefore, $ f(x) $ is an odd function.