If `f(x)={{:((sqrt(x^(2)+7)-4)/(x+3)",",xne-3),(k,","x=-3):}` is continuous at x=-3 then the value of k is

Options

A. `(3)/(4)`
B. 0
C. `-(3)/(4)`
D. none of these

Answer

To determine the value of \( k \) such that the function

\[ f(x) = \begin{cases} \frac{\sqrt{x^2 + 7} - 4}{x + 3} & \text{if } x \neq -3 \\ k & \text{if } x = -3 \end{cases} \]

is continuous at \( x = -3 \), we need to ensure that the limit of \( f(x) \) as \( x \) approaches \(-3\) is equal to \( f(-3) \).

### Step 1: Find the limit of \( f(x) \) as \( x \) approaches \(-3 \)

We start by calculating the limit:

\[ \lim_{x \to -3} f(x) = \lim_{x \to -3} \frac{\sqrt{x^2 + 7} - 4}{x + 3} \]

### Step 2: Substitute \( x = -3 \) directly into the function

Substituting \( x = -3 \) gives:

\[ f(-3) = \frac{\sqrt{(-3)^2 + 7} - 4}{-3 + 3} = \frac{\sqrt{9 + 7} - 4}{0} = \frac{\sqrt{16} - 4}{0} = \frac{4 - 4}{0} = \frac{0}{0} \]

This is an indeterminate form, so we need to apply L'Hôpital's Rule.

### Step 3: Apply L'Hôpital's Rule

According to L'Hôpital's Rule, we differentiate the numerator and denominator:

1. Differentiate the numerator \( \sqrt{x^2 + 7} - 4 \):
- The derivative of \( \sqrt{x^2 + 7} \) is \( \frac{1}{2\sqrt{x^2 + 7}} \cdot (2x) = \frac{x}{\sqrt{x^2 + 7}} \).
- The derivative of \(-4\) is \(0\).

Thus, the derivative of the numerator is:

\[ \frac{x}{\sqrt{x^2 + 7}} \]

2. Differentiate the denominator \( x + 3 \):
- The derivative is \(1\).

Now we can apply L'Hôpital's Rule:

\[ \lim_{x \to -3} \frac{\sqrt{x^2 + 7} - 4}{x + 3} = \lim_{x \to -3} \frac{\frac{x}{\sqrt{x^2 + 7}}}{1} = \lim_{x \to -3} \frac{x}{\sqrt{x^2 + 7}} \]

### Step 4: Evaluate the limit

Now substituting \( x = -3 \):

\[ \lim_{x \to -3} \frac{x}{\sqrt{x^2 + 7}} = \frac{-3}{\sqrt{(-3)^2 + 7}} = \frac{-3}{\sqrt{9 + 7}} = \frac{-3}{\sqrt{16}} = \frac{-3}{4} \]

### Step 5: Set the limit equal to \( k \)

For the function to be continuous at \( x = -3 \):

\[ k = \lim_{x \to -3} f(x) = \frac{-3}{4} \]

### Conclusion

Thus, the value of \( k \) is

\[ \boxed{-\frac{3}{4}} \]