For what values of constant K, the following functions are continuous the indicated points.
`f(x)={{:(,(1-cos 4x)/(x^(2)), x lt 0),(,K,x=0),(,sqrt(x)+8,x gt 0):}" at x =0"`

Answer

To determine the values of the constant \( K \) for which the function

\[ f(x) = \begin{cases} \frac{1 - \cos(4x)}{x^2} & \text{if } x < 0 \\ K & \text{if } x = 0 \\ \sqrt{x} + 8 & \text{if } x > 0 \end{cases} \]

is continuous at \( x = 0 \), we need to ensure that the left-hand limit and right-hand limit at \( x = 0 \) are equal to the value of the function at that point, which is \( K \).

### Step 1: Calculate the Left-Hand Limit as \( x \to 0^- \)

The left-hand limit is given by:

\[ \lim_{x \to 0^-} f(x) = \lim_{x \to 0^-} \frac{1 - \cos(4x)}{x^2} \]

### Step 2: Evaluate the Limit

As \( x \to 0 \), both the numerator and denominator approach 0, leading to an indeterminate form \( \frac{0}{0} \). We can apply L'Hôpital's Rule:

1. Differentiate the numerator: The derivative of \( 1 - \cos(4x) \) is \( 4\sin(4x) \).
2. Differentiate the denominator: The derivative of \( x^2 \) is \( 2x \).

Thus, we have:

\[ \lim_{x \to 0^-} \frac{1 - \cos(4x)}{x^2} = \lim_{x \to 0^-} \frac{4\sin(4x)}{2x} = 2 \lim_{x \to 0^-} \frac{\sin(4x)}{x} \]

### Step 3: Simplify the Limit

We can rewrite the limit as:

\[ 2 \lim_{x \to 0^-} \frac{\sin(4x)}{4x} \cdot 4 = 2 \cdot 1 \cdot 4 = 8 \]

So, the left-hand limit is:

\[ \lim_{x \to 0^-} f(x) = 8 \]

### Step 4: Calculate the Right-Hand Limit as \( x \to 0^+ \)

The right-hand limit is given by:

\[ \lim_{x \to 0^+} f(x) = \lim_{x \to 0^+} (\sqrt{x} + 8) \]

As \( x \to 0 \):

\[ \sqrt{x} \to 0 \quad \text{and thus} \quad \lim_{x \to 0^+} f(x) = 0 + 8 = 8 \]

### Step 5: Set the Limits Equal to Each Other

For the function to be continuous at \( x = 0 \), we need:

\[ \lim_{x \to 0^-} f(x) = \lim_{x \to 0^+} f(x) = f(0) \]

This gives us:

\[ 8 = K \]

### Conclusion

The value of the constant \( K \) for which the function is continuous at \( x = 0 \) is:

\[ K = 8 \]