Even function?

Question

Why ((x-1)^2)^(1/3)+((x+1)^2)^(1/3) is an even function?

Answer 1

For a function to be "even", f(-x) must equal f(x) for all x in the domain of the function.  

 

If f(x) = [(x-1)²]^1/3 + [(x+1)²]^1/3

 

          = (x-1)^2/3 + (x+1)^2/3

 

then f(-x) = (-x-1)^2/3 + (-x+1)^2/3

 

               = [-1(x+1)]^2/3 + [-1(x-1)]^2/3

 

               = (-1)^2/3(x+1)^2/3 + (-1)^2/3(x-1)^2/3   NOTE:  (-1)^2/3 = (³√(-1) )² = 1

 

               = (x+1)^2/3 + (x-1)^2/3

             

               = f(x)

 

So, f is an even function.