If `f(x)=x^(2)+kx+1` for all x and f is an even function, find `k,kinR`.

Answer

To solve the problem, we need to find the value of \( k \) such that the function \( f(x) = x^2 + kx + 1 \) is an even function. An even function satisfies the condition \( f(-x) = f(x) \) for all \( x \).

### Step-by-step Solution:

1. **Write down the function**:
\[ f(x) = x^2 + kx + 1 \]

2. **Determine \( f(-x) \)**:
Substitute \(-x\) into the function:
\[ f(-x) = (-x)^2 + k(-x) + 1 \]
Simplifying this gives:
\[ f(-x) = x^2 - kx + 1 \]

3. **Set up the equation for even function**:
Since \( f \) is an even function, we have:
\[ f(-x) = f(x) \]
Thus, we can set the two expressions equal to each other:
\[ x^2 - kx + 1 = x^2 + kx + 1 \]

4. **Simplify the equation**:
Cancel out \( x^2 \) and \( 1 \) from both sides:
\[ -kx = kx \]

5. **Combine like terms**:
This simplifies to:
\[ -kx - kx = 0 \]
or
\[ -2kx = 0 \]

6. **Solve for \( k \)**:
Since this must hold for all \( x \), the only way for this equation to be true is if:
\[ k = 0 \]

7. **Conclusion**:
Therefore, the value of \( k \) such that \( f(x) \) is an even function is:
\[ k = 0 \]