SOLUTION: Determine whether function;f(x) =(-1)^[x] is even,odd or neither of two (where[]denotes the greatest integer function).

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Question 1196650: Determine whether function;f(x) =(-1)^[x] is even,odd or neither of two (where[]denotes the greatest integer function).
Found 2 solutions by ikleyn, MathLover1:
Answer by ikleyn(52754) About Me  (Show Source):
You can put this solution on YOUR website!
.
Determine whether function f(x) =(-1)^[x] is even, odd or neither of two
(where [] denotes the greatest integer function).
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DEFINITION: Greatest Integer Function

    The greatest integer function is also known as the step function. It RETURNS the number which is the nearest integer 
    less than or equal to the given number. The greatest integer function has a step curve 
    The domain of the greatest integer function is ℝ and its range is ℤ.

    Therefore the greatest integer function is simply rounding off the given number to the greatest 
    integer that is less than or equal to the given number. 


See the link https://www.cuemath.com/algebra/greatest-integer-function

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Let's consider our function at x= 1.2.

Then  [x] = [1.2] = 1.  THEREFPRE, f(x) = %28-1%29%5E1 = -1.



At x= -1.2, we have [x] = [-1.2] = -2.  THEREFORE, f(x) = f(-1.2) = %28-1%29%5E%28-2%29 = 1.

Since f(1.2) =/= f(-1.2), f(x) is not an even function.




Let's consider our function at x= 2.

Then  [x] = [2] = 2.  THEREFPRE, f(x) = %28-1%29%5E2 = 1.



At x= -2, we have [x] = [-2] = -2.  THEREFORE, f(x) = f(-2) = %28-1%29%5E%28-2%29 = 1.

Since f(-2) =/= -f(-2), f(x) is not an odd function.



So, the conclusion is that given function f(x) is  NEITHER  even  NOR  odd.

Completed, proved and explained.


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The logic and the reasoning in the post by @MathLover1 are defective and lead you to NOWHERE.

So, you better ignore her post, for your safety.


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This problem is tricky: on the set of integer numbers, function f(x) is EVEN.
It is because for integer x, the number x is always the same as [x], so [x] and [-x] always have the same parity.

On the set of all other real numbers (others than integers), function f(x) is ODD.
It is because for non-integer x, the number [x] always has different parity than [-x].


Considered on the total set of all real numbers, function f(x) is NEITHER even NOR odd, as I said it in my post.

The cause for it is that the function [x] itself is NEITHER even NOR odd on the set of all real numbers.



Answer by MathLover1(20849) About Me  (Show Source):
You can put this solution on YOUR website!
f%28x%29+=%28-1%29^[x] is even, odd or neither
Let f:RR be a function defined as f%28x%29=%28-1%29^[x]
.
When x+is an integer [x]=x, hence f%28x%29=%28-1%29^[x] =%28-1%29%5Ex, which is an even function.

If x is a positive real number which is not an integer, then [x] is the integer part of the given number, say a.
Therefore f%28x%291 based on a is even or odd.
If x+is a negative real number which is not an integer, then [x]=a%5E-1, where a is the integer part of the given number.
Therefore f%28x%291 based on a is odd or even.
Hence f%28x%29=-f%28-x%29 when x not ∈Z+

so, correct answers are:
f%28x%29 is an even function when+x+Z
f%28x%29 is an odd+function when x is notZ
f%28x%29 is neither+even nor odd