If `f(x) = 3(2x + 3)^(2//3) + 2x + 3`, then:

Options

A. (a) `f(x)` is continuous but not differentiable at `x = - (3)/(2)`
B. (b) `f(x)` is differentiable at `x = 0`
C. (c) `f(x)` is continuous at `x = 0`
D. (d) `f(x)` is differentiable but not continuous at `x = - (3)/(2)`

Answer

To solve the problem, we need to analyze the function \( f(x) = 3(2x + 3)^{\frac{2}{3}} + 2x + 3 \) for its continuity and differentiability at the points \( x = 0 \) and \( x = -\frac{3}{2} \).

### Step 1: Check Continuity at \( x = 0 \)

To check continuity at \( x = 0 \), we need to find \( f(0) \) and ensure that the limit as \( x \) approaches 0 equals \( f(0) \).

1. Calculate \( f(0) \):
\[ f(0) = 3(2(0) + 3)^{\frac{2}{3}} + 2(0) + 3 = 3(3)^{\frac{2}{3}} + 3 = 3 \cdot 3^{\frac{2}{3}} + 3 \]
Since \( 3^{\frac{2}{3}} = \sqrt[3]{9} \), we can simplify:
\[ f(0) = 3 \cdot \sqrt[3]{9} + 3 \]

2. Calculate the limit as \( x \) approaches 0:
\[ \lim_{x \to 0} f(x) = \lim_{x \to 0} \left( 3(2x + 3)^{\frac{2}{3}} + 2x + 3 \right) \]
Substitute \( x = 0 \):
\[ = 3(3)^{\frac{2}{3}} + 3 = 3 \cdot \sqrt[3]{9} + 3 \]
Thus, \( \lim_{x \to 0} f(x) = f(0) \).

Since both the limit and the function value at \( x = 0 \) are equal, \( f \) is continuous at \( x = 0 \).

### Step 2: Check Differentiability at \( x = 0 \)

To check differentiability at \( x = 0 \), we need to find \( f'(x) \).

1. Differentiate \( f(x) \):
\[ f'(x) = 3 \cdot \frac{2}{3}(2x + 3)^{-\frac{1}{3}} \cdot 2 + 2 \]
Simplifying this:
\[ f'(x) = 2(2x + 3)^{-\frac{1}{3}} \cdot 2 + 2 = \frac{4}{(2x + 3)^{\frac{1}{3}}} + 2 \]

2. Evaluate \( f'(0) \):
\[ f'(0) = \frac{4}{(2(0) + 3)^{\frac{1}{3}}} + 2 = \frac{4}{3^{\frac{1}{3}}} + 2 \]
This is defined, so \( f \) is differentiable at \( x = 0 \).

### Step 3: Check Continuity and Differentiability at \( x = -\frac{3}{2} \)

1. Calculate \( f\left(-\frac{3}{2}\right) \):
\[ f\left(-\frac{3}{2}\right) = 3(2(-\frac{3}{2}) + 3)^{\frac{2}{3}} + 2(-\frac{3}{2}) + 3 = 3(0)^{\frac{2}{3}} - 3 + 3 = 0 \]

2. Check the limit as \( x \) approaches \( -\frac{3}{2} \):
\[ \lim_{x \to -\frac{3}{2}} f(x) = f\left(-\frac{3}{2}\right) = 0 \]
Thus, \( f \) is continuous at \( x = -\frac{3}{2} \).

3. Check differentiability at \( x = -\frac{3}{2} \):
\[ f'(-\frac{3}{2}) = \frac{4}{(2(-\frac{3}{2}) + 3)^{\frac{1}{3}}} + 2 = \frac{4}{0^{\frac{1}{3}}} + 2 \]
This is undefined, so \( f \) is not differentiable at \( x = -\frac{3}{2} \).

### Conclusion

- \( f \) is continuous at \( x = 0 \) and differentiable at \( x = 0 \).
- \( f \) is continuous at \( x = -\frac{3}{2} \) but not differentiable.

### Final Answers:
- a) True: \( f \) is continuous but not differentiable at \( -\frac{3}{2} \).
- b) True: \( f \) is differentiable at \( 0 \).
- c) True: \( f \) is continuous at \( 0 \).
- d) False: \( f \) is not differentiable but continuous at \( -\frac{3}{2} \).