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To ensure the function \( f(x) \) is continuous everywhere, the two pieces of the function must be equal at the point where they meet, which is at \( x = \frac{5\pi}{6} \). Let's set the two expressions for \( f(x) \) equal to each other at \( x = \frac{5\pi}{6} \).

Given:
\[
f(x) = 
\begin{cases}
6 \sin(x) + 2 & \text{if } x \leq \frac{5\pi}{6} \\
mx + 2 & \text{if } x \geq \frac{5\pi}{6}
\end{cases}
\]

For continuity at \( x = \frac{5\pi}{6} \), we need:

\[
6 \sin\left(\frac{5\pi}{6}\right) + 2 = m \left(\frac{5\pi}{6}\right) + 2.
\]

Let's calculate \( \sin\left(\frac{5\pi}{6}\right) \) and solve for \( m \).

\[
\sin\left(\frac{5\pi}{6}\right) = \sin\left(\pi - \frac{\pi}{6}\right) = \sin\left(\frac{\pi}{6}\right) = \frac{1}{2}.
\]

Substitute this value into the equation:

\[
6 \cdot \frac{1}{2} + 2 = m \left(\frac{5\pi}{6}\right) + 2.
\]

Simplify the equation:

\[
3 + 2 = m \left(\frac{5\pi}{6}\right) + 2.
\]

\[
5 = m \left(\frac{5\pi}{6}\right).
\]

Solve for \( m \):

\[
m = \frac{5 \cdot 6}{5\pi} = \frac{6}{\pi}.
\]

Therefore, the value of \( m \) that makes the function \( f(x) \) continuous everywhere is:

\[
\boxed{\frac{6}{\pi}}.
\]

Would you like further details or have any questions?

### Related Questions:
1. How do you prove a function is differentiable?
2. What are some properties of continuous functions?
3. How do we handle piecewise functions in integration?
4. What are some other examples of functions where continuity needs to be checked?
5. How can we generalize continuity for multivariable functions?

**Tip:** Remember, checking continuity at a point involves verifying both the limit and the function value at that point!

Let f(x) = {6sin(x) + 2 if x <= 5π/6; mx + 2 if x >= 5π/6}. What must the value of m be so that the function f(x) is continuous everywhere? Give result in exact form.

Learn how to determine the value of m to ensure the piecewise function f(x) is continuous at x = 5π/6. This example covers key calculus concepts like continuity, limits, and trigonometric functions.

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