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To determine the values of \( k \) and \( p \) such that the function \( f(x) \) is both continuous and differentiable at \( x = 1 \), we need to ensure that:

1. **Continuity**: The left-hand limit \( \lim_{x \to 1^-} f(x) \) and right-hand limit \( \lim_{x \to 1^+} f(x) \) must be equal to \( f(1) \).
2. **Differentiability**: The derivatives from the left and right at \( x = 1 \) must also be equal.

The function \( f(x) \) is defined as:
\[
f(x) = \begin{cases} 
2 - x^2, & x \leq 1 \\
kx + p, & x > 1 
\end{cases}
\]

### Step 1: Continuity at \( x = 1 \)
To ensure continuity, we need:
\[
\lim_{x \to 1^-} f(x) = \lim_{x \to 1^+} f(x) = f(1)
\]

#### Left-hand limit as \( x \to 1^- \):
\[
\lim_{x \to 1^-} f(x) = 2 - (1)^2 = 2 - 1 = 1
\]

#### Right-hand limit as \( x \to 1^+ \):
For \( x > 1 \), \( f(x) = kx + p \). So:
\[
\lim_{x \to 1^+} f(x) = k(1) + p = k + p
\]

To satisfy continuity, these limits must be equal:
\[
1 = k + p
\]

### Step 2: Differentiability at \( x = 1 \)
To ensure differentiability, we also need the derivatives from the left and right at \( x = 1 \) to be equal.

#### Derivative from the left (using \( f(x) = 2 - x^2 \) for \( x \leq 1 \)):
\[
f'(x) = -2x
\]
Thus,
\[
\lim_{x \to 1^-} f'(x) = -2(1) = -2
\]

#### Derivative from the right (using \( f(x) = kx + p \) for \( x > 1 \)):
\[
f'(x) = k
\]
Thus,
\[
\lim_{x \to 1^+} f'(x) = k
\]

For differentiability, we need:
\[
-2 = k
\]

### Solving for \( k \) and \( p \)
From continuity, we found:
\[
1 = k + p
\]
Substituting \( k = -2 \) into this equation:
\[
1 = -2 + p
\]
\[
p = 3
\]

### Conclusion
The values of \( k \) and \( p \) that make \( f(x) \) both continuous and differentiable at \( x = 1 \) are:
\[
k = -2 \quad \text{and} \quad p = 3
\]

Would you like further details or have any questions?

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Here are five related questions to consider:
1. What if the function were continuous but not differentiable at \( x = 1 \)? How would that affect \( k \) and \( p \)?
2. Can you verify the differentiability of \( f(x) \) at other points, such as \( x = 0 \)?
3. What happens if \( k \) or \( p \) take values different from those we calculated?
4. How would the analysis change if the function were defined differently for \( x < 1 \)?
5. Can we generalize this approach for other piecewise functions with quadratic and linear parts?

**Tip:** For any piecewise function, always start by checking continuity before differentiability at the point where the definition changes.

For what values of k and p is the function f(x) both continuous and differentiable at x = 1?

Learn how to find the values of constants k and p such that a piecewise function is both continuous and differentiable at x = 1. This problem covers important calculus concepts like limits, continuity, and differentiability. Ideal for high school calculus students.

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