How do you determine if #x^3-x^2-x+1# is an even or odd function?

Question

Ariana Alvarez · Accepted Answer

This function is neither even nor odd.

#f(x)# is an even function if #f(-x) = f(x)# for all #x#. #f(x)# is an odd function if #f(-x) = -f(x)# for all #x#. Let #f(x) = x^3-x^2-x+1# Then: #f(2) = 8-4-2+1 = 3# #f(-2) = -8+4+2+1 = -1# So #f(-2) != f(2)# and #f(-2) != -f(2)# So this #f(x)# is neither even nor odd. #color(white)()# Actually, since this is a polynomial, we could tell by looking at the degrees of the terms: A polynomial is an even function if and only if all of its terms are of even degree. Note that a constant term is of even degree since it has degree #0#. A polynomial is an odd function if and only if all of its terms are of odd degree. #color(white)()# The same is true of power series: A power series represents an even function if and only if all of its terms are of even degree. For example, #cos x = sum_(n=0)^oo (-1)^n x^(2n)/((2n)!)# is an even function. A power series represents an odd function if and only if all of its terms are of odd degree. For example, #sin x = sum_(n=0)^oo (-1)^n x^(2n+1)/((2n+1)!)# is an odd function.

Greyson Brown · Answer

undefined To determine if the function $ f(x) = x^3 - x^2 - x + 1 $ is even or odd, we evaluate the function for $ f(-x) $ and compare it with $ f(x) $.

If $ f(-x) = f(x) $, the function is even.

If $ f(-x) = -f(x) $, the function is odd.

For the given function:

$ f(-x) = (-x)^3 - (-x)^2 - (-x) + 1 $

$ = -x^3 - x^2 + x + 1 $

Comparing with $ f(x) = x^3 - x^2 - x + 1 $:

$ f(-x) \neq f(x) $

$ f(-x) $ is not equal to $ f(x) $ and not equal to the negation of $ f(x) $, so the function is neither even nor odd.

Chloe Alexander · Answer

undefined To determine if $x^3 - x^2 - x + 1$ is an even or odd function, follow these steps:

1. Substitute $-x$ for $x$ in the function.
2. Simplify the expression.
3. Compare the result with the original function.

If the result is equal to the original function, then the function is even. If the result is equal to the negative of the original function, then the function is odd. If neither condition is met, the function is neither even nor odd.

Let's apply this to $x^3 - x^2 - x + 1$:

1. Substitute $-x$ for $x$:
$$(-x)^3 - (-x)^2 - (-x) + 1$$

2. Simplify the expression:
$$-x^3 - x^2 + x + 1$$

3. Compare with the original function:
$$x^3 - x^2 - x + 1$$

Since the result, $-x^3 - x^2 + x + 1$, is not equal to either the original function or the negative of the original function, $x^3 - x^2 - x + 1$ is neither an even nor an odd function.

Caleb Alexander · Answer

undefined To determine if $ x^3 - x^2 - x + 1 $ is an even or odd function, we need to examine its symmetry properties.

1. Even Function:
   - If $ f(-x) = f(x) $ for all $ x $ in the domain, the function is even.
   - Substitute $ -x $ into the function and simplify. If the result is equivalent to the original function, it's even.

2. Odd Function:
   - If $ f(-x) = -f(x) $ for all $ x $ in the domain, the function is odd.
   - Substitute $ -x $ into the function and simplify. If the result is the negative of the original function, it's odd.

For $ f(x) = x^3 - x^2 - x + 1 $:

1. Even:
   - $ f(-x) = (-x)^3 - (-x)^2 - (-x) + 1 $
   - $ f(-x) = -x^3 - x^2 + x + 1 $

This is not equivalent to the original function $ f(x) $, so it's not even.

2. Odd:
   - $ f(-x) = -x^3 - x^2 + x + 1 $
   - $ -f(x) = -(x^3 - x^2 - x + 1) = -x^3 + x^2 + x - 1 $

This is also not equivalent to $ f(-x) $, nor is it the negative of $ f(x) $, so it's not odd.

Conclusion:
$ x^3 - x^2 - x + 1 $ is neither an even nor an odd function.