To solve this problem, we need to use the fact that the function $f$ is continuous at $x = 6$. Since $f$ is continuous, the left-hand limit and the right-hand limit at $x = 6$ must be equal.

Given:

1. For $x < 6$: $f(x) = \cos(x)$.
2. For $x \geq 6$: $f(x) = x + a$.

To ensure continuity at $x = 6$, we set up the following equation:

$\lim_{x \to 6^-} f(x) = \lim_{x \to 6^+} f(x)$

Step 1: Calculate the left-hand limit at $x = 6$

For $x < 6$, $f(x) = \cos(x)$. So, the left-hand limit is: $\lim_{x \to 6^-} f(x) = \cos(6)$

Step 2: Calculate the right-hand limit at $x = 6$

For $x \geq 6$, $f(x) = x + a$. So, the right-hand limit is: $\lim_{x \to 6^+} f(x) = 6 + a$

Step 3: Set the limits equal to ensure continuity

Since $f$ is continuous at $x = 6$, we equate the left-hand and right-hand limits: $\cos(6) = 6 + a$

Step 4: Solve for $a$

Rearrange the equation to solve for $a$: $a = \cos(6) - 6$

Now, calculating the numerical value of $\cos(6)$ (in radians): $\cos(6) \approx 0.9602$ $a \approx 0.9602 - 6 = -5.0398$

Final Answer:

$a \approx -5.04$

Would you like additional details, or do you have any questions?

Related Questions

1. What does continuity mean for a function at a specific point?
2. How can we calculate limits from the left-hand side and right-hand side?
3. How does one determine if a piecewise function is continuous?
4. What are the general properties of the cosine function?
5. How can we approximate trigonometric values like $\cos(6)$ without a calculator?

Tip

For piecewise functions, always ensure continuity by matching the values from the left and right at the boundaries.