Question 17

f(x) could NOT be (x^3-1)^2

The function f(x) cannot be (x^3-1)^2 because it is not an even function, which is required for the condition f(x) = f(-x) to hold for all x. This function has terms that result in asymmetry when evaluated at -x.

A) x²-2

This function is even because f(-x) = (-x)² - 2 = x² - 2, which equals f(x). Thus, it satisfies the condition f(x) = f(-x).

B) x²+2

This function is also even since f(-x) = (-x)² + 2 = x² + 2, which is equal to f(x). Hence, it meets the requirement of being symmetric.

C) x^4-x²

This function is not even because f(-x) = (-x)⁴ - (-x)² = x⁴ - x², which equals f(x). Therefore, it satisfies the condition f(x) = f(-x).

D) (x^3-1)^2

This function is not even because f(-x) = ((-x)³ - 1)² = (-x^3 - 1)², which does not equal f(x). The asymmetry in the cubic term leads to a different value when evaluated at -x, violating the even function requirement.

E) (x^3-x)^2

This function is even because f(-x) = ((-x)³ - (-x))² = (-x^3 + x)² = (x^3 - x)², which equals f(x). Thus, it meets the condition of being symmetric.

Conclusion

The function (x^3-1)^2 is definitively not even, as it fails to satisfy the condition f(x) = f(-x) due to the inherent asymmetry of the cubic term. In contrast, the other options either are even functions or fulfill the required symmetry, making them valid candidates for f(x).