unction is even, odd, or neither. \( f(x)=(x+1)^{2} \)

To determine if the function f(x) = (x + 1)² is even, odd, or neither, we start by recalling the definitions: • A function is even if f(-x) = f(x) for every x. • A function is odd if f(-x) = -f(x) for every x. Step 1: Compute f(-x)   f(-x) = ((-x) + 1)² = (1 - x)² Step 2: Test the even property We need to see if (1 - x)² = (x + 1)² for every x. Comparing these two expressions:   For x = 1:    f(1) = (1 + 1)² = 4,    f(-1) = (1 - 1)² = 0. Since f(1) ≠ f(-1), the function is not even. Step 3: Test the odd property We now need to see if (1 - x)² = - (x + 1)² for every x.   Again using x = 1:    f(1) = 4,    - f(1) = -4, but f(-1) = 0. This shows that f(-x) ≠ -f(x), so the function is not odd. Conclusion: Since f(x) = (x + 1)² does not satisfy the criteria for being even or odd, the function is neither even nor odd.