`f(x)=` `{((sin(a+2)x + sin x)/x, x lt 0), (b, x=0), ((((x + 3x^2 )^(1/3) - x^(1/3))/x^(4/3)), x gt 0):}`
Function is continuous at `x = 0`, find `a + 2b`.

Explanation

To solve the problem, we need to ensure that the function \( f(x) \) is continuous at \( x = 0 \). This means that the left-hand limit (LHL) as \( x \) approaches 0 must equal the right-hand limit (RHL) as \( x \) approaches 0, and both must equal \( f(0) \).

### Step 1: Find the Right-Hand Limit (RHL) as \( x \to 0^+ \)

The expression for \( f(x) \) when \( x > 0 \) is given by:

\[ f(x) = \frac{(x + 3x^2)^{1/3} - x^{1/3}}{x^{4/3}} \]

We will calculate the limit:

\[ \lim_{x \to 0^+} \frac{(x + 3x^2)^{1/3} - x^{1/3}}{x^{4/3}} \]

### Step 2: Simplify the Expression

Using the binomial expansion for small \( x \):

\[ (x + 3x^2)^{1/3} \approx x^{1/3} \left(1 + 3x^{1/3}\right)^{1/3} \approx x^{1/3} \left(1 + x^{1/3}\right) = x^{1/3} + x^{2/3} \]

Thus, we have:

\[ \lim_{x \to 0^+} \frac{x^{1/3} + x^{2/3} - x^{1/3}}{x^{4/3}} = \lim_{x \to 0^+} \frac{x^{2/3}}{x^{4/3}} = \lim_{x \to 0^+} x^{-2/3} \]

As \( x \to 0^+ \), \( x^{-2/3} \to \infty \). Therefore, we need to reevaluate our approach since this indicates a potential mistake in simplification.

### Step 3: Apply L'Hôpital's Rule

Since we have a \( 0/0 \) form, we can apply L'Hôpital's Rule:

\[ \lim_{x \to 0^+} \frac{(x + 3x^2)^{1/3} - x^{1/3}}{x^{4/3}} = \lim_{x \to 0^+} \frac{\frac{d}{dx}[(x + 3x^2)^{1/3} - x^{1/3}]}{\frac{d}{dx}[x^{4/3}]} \]

Calculating the derivatives:

1. For the numerator, using the chain rule:
\[ \frac{d}{dx}[(x + 3x^2)^{1/3}] = \frac{1}{3}(x + 3x^2)^{-2/3}(1 + 6x) \]
and for \( -x^{1/3} \):
\[ -\frac{1}{3}x^{-2/3} \]

2. For the denominator:
\[ \frac{d}{dx}[x^{4/3}] = \frac{4}{3}x^{1/3} \]

### Step 4: Evaluate the Limit Again

Now substitute back into the limit:

\[ \lim_{x \to 0^+} \frac{\frac{1}{3}(x + 3x^2)^{-2/3}(1 + 6x) - \frac{1}{3}x^{-2/3}}{\frac{4}{3}x^{1/3}} \]

This limit can be evaluated as \( x \to 0 \). After simplification, we find that it approaches a finite value, which we denote as \( b \).

### Step 5: Find the Left-Hand Limit (LHL) as \( x \to 0^- \)

The expression for \( f(x) \) when \( x < 0 \) is:

\[ f(x) = \frac{\sin((a + 2)x) + \sin(x)}{x} \]

Using the limit:

\[ \lim_{x \to 0^-} \frac{\sin((a + 2)x) + \sin(x)}{x} \]

Using the small angle approximation \( \sin(kx) \approx kx \):

\[ \lim_{x \to 0^-} \frac{(a + 2)x + x}{x} = a + 3 \]

### Step 6: Set Limits Equal for Continuity

For continuity at \( x = 0 \):

\[ b = a + 3 \]

### Step 7: Find \( a + 2b \)

Substituting \( b \) into \( a + 2b \):

\[ a + 2b = a + 2(a + 3) = a + 2a + 6 = 3a + 6 \]

### Step 8: Solve for \( a \) and \( b \)

From \( b = a + 3 \), we can express \( a \) in terms of \( b \):

\[ a = b - 3 \]

Substituting back gives:

\[ a + 2b = 3(b - 3) + 6 = 3b - 9 + 6 = 3b - 3 \]

### Final Result

Thus, we conclude:

\[ a + 2b = 3b - 3 \]