To ensure that $f(x)$ is continuous and differentiable at $x = 0$, we must make sure that both $f(x)$ and $f'(x)$ match at $x = 0$ from the left and the right.

Step 1: Continuity at $x = 0$

For $f(x)$ to be continuous at $x = 0$: $\lim_{x \to 0^-} f(x) = \lim_{x \to 0^+} f(x) = f(0).$

Left-hand limit ($x \to 0^-$): Since $f(x) = 3x + 5$ when $x < 0$, $\lim_{x \to 0^-} f(x) = 3(0) + 5 = 5.$

Right-hand limit ($x \to 0^+$): Since $f(x) = \sin(6ax) + b$ when $x > 0$, $\lim_{x \to 0^+} f(x) = \sin(0) + b = b.$

For continuity, these limits must be equal: $5 = b.$

Thus, $b = 5$.

Step 2: Differentiability at $x = 0$

For $f'(0)$ to exist, the left-hand and right-hand derivatives at $x = 0$ must be equal.

Left-hand derivative ($x < 0$): $f(x) = 3x + 5 \Rightarrow f'(x) = 3.$ So, $\lim_{x \to 0^-} f'(x) = 3.$

Right-hand derivative ($x > 0$): Since $f(x) = \sin(6ax) + 5$, we differentiate $f(x)$ with respect to $x$: $f'(x) = 6a \cos(6ax).$ Thus, $\lim_{x \to 0^+} f'(x) = 6a \cos(0) = 6a.$

For $f'(0)$ to exist, we must have: $3 = 6a.$ Solving for $a$: $a = \frac{1}{2}.$

Conclusion

The values of $a$ and $b$ that make $f(x)$ continuous and differentiable at $x = 0$ are: $a = \frac{1}{2}, \quad b = 5.$

Would you like a more detailed explanation, or do you have any questions?

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Related Questions:

1. What is the role of continuity and differentiability in defining a function's smoothness?
2. How would the solution change if $f(x)$ had a different expression for $x < 0$?
3. What happens if only continuity is required and not differentiability?
4. How does differentiability at a point imply continuity, but not vice versa?
5. Could there be any other values for $a$ and $b$ if different conditions were given?

Tip: For functions with piecewise definitions, always start by checking continuity and then differentiability at the boundaries.