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

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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.
The function $$ f(x) = (x + 1)^2 $$ is neither even nor odd.

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

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