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We are asked to explain why there is no value of \( k \) that will make the following piecewise function continuous:

\[
f(x) =
\begin{cases}
5x & \text{if } x < -1 \\
x + k & \text{if } -1 \leq x \leq 3 \\
2x^2 & \text{if } x > 3
\end{cases}
\]

For a function to be continuous at a given point, it must satisfy three conditions:
1. The function must be defined at the point.
2. The limit of the function as \( x \) approaches the point from both the left and the right must exist.
3. The value of the function at the point must equal the limit.

We need to check the potential discontinuities at the boundaries where the function changes, i.e., at \( x = -1 \) and \( x = 3 \).

### 1. Continuity at \( x = -1 \)

- For \( x < -1 \), the function is \( f(x) = 5x \). So, as \( x \to -1^- \):
  \[
  \lim_{x \to -1^-} f(x) = 5(-1) = -5.
  \]

- For \( -1 \leq x \leq 3 \), the function is \( f(x) = x + k \). So, at \( x = -1 \):
  \[
  f(-1) = -1 + k.
  \]

- For the function to be continuous at \( x = -1 \), we require:
  \[
  \lim_{x \to -1^-} f(x) = f(-1).
  \]
  This gives:
  \[
  -5 = -1 + k \quad \Rightarrow \quad k = -4.
  \]

So, continuity at \( x = -1 \) can be achieved if \( k = -4 \).

### 2. Continuity at \( x = 3 \)

- For \( -1 \leq x \leq 3 \), the function is \( f(x) = x + k \). So, as \( x \to 3^- \):
  \[
  \lim_{x \to 3^-} f(x) = 3 + k.
  \]

- For \( x > 3 \), the function is \( f(x) = 2x^2 \). So, as \( x \to 3^+ \):
  \[
  \lim_{x \to 3^+} f(x) = 2(3^2) = 18.
  \]

- For the function to be continuous at \( x = 3 \), we require:
  \[
  \lim_{x \to 3^-} f(x) = \lim_{x \to 3^+} f(x).
  \]
  This gives:
  \[
  3 + k = 18 \quad \Rightarrow \quad k = 15.
  \]

### Conclusion:
To make the function continuous at \( x = -1 \), we need \( k = -4 \), but to make it continuous at \( x = 3 \), we need \( k = 15 \). Since \( k \) cannot take two different values simultaneously, there is no single value of \( k \) that will make the function continuous at both \( x = -1 \) and \( x = 3 \). Therefore, there is no value of \( k \) that will make the function continuous everywhere.

Would you like more details or explanations on any step?

Here are some related questions for deeper exploration:
1. What is the general condition for a piecewise function to be continuous?
2. How would you modify the function to make it continuous at both points?
3. Can a function be continuous on some intervals but not at specific points?
4. What is the difference between a jump discontinuity and a removable discontinuity?
5. How does the continuity of a quadratic function compare to that of a linear function?

**Tip:** Continuity often involves ensuring that the limits from both sides of a boundary point match the function's value at that point. This can be especially tricky in piecewise functions!

Explain why there is no value of k that will make the following function continuous:

f(x) =
  5x, if x < -1
  x + k, if -1 <= x <= 3
  2x^2, if x > 3

This problem involves a piecewise function where students must prove that no value of k makes the function continuous across its defined intervals. It covers the concept of limits and continuity, with focus points at x = -1 and x = 3.

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