Let `f(x)=(3x-7)x^(2/3)`. The interval in which f(x) is increasing.

Options

A. `(0,14/15)`
B. `(-oo,0)uu(14/15,oo)`
C. `(-oo,14/15)`
D. `(0,oo)`

Answer

To determine the interval in which the function \( f(x) = (3x - 7)x^{2/3} \) is increasing, we will follow these steps:

### Step 1: Find the derivative \( f'(x) \)

Using the product rule, where \( u = 3x - 7 \) and \( v = x^{2/3} \):
\[ f'(x) = u'v + uv' \]
Calculating \( u' \) and \( v' \):
- \( u' = 3 \)
- \( v' = \frac{2}{3}x^{-1/3} \)

Now substituting back into the product rule:
\[ f'(x) = (3)(x^{2/3}) + (3x - 7)\left(\frac{2}{3}x^{-1/3}\right) \]
\[ f'(x) = 3x^{2/3} + \frac{2(3x - 7)}{3}x^{-1/3} \]

### Step 2: Simplify \( f'(x) \)

To simplify \( f'(x) \):
\[ f'(x) = 3x^{2/3} + \frac{2(3x - 7)}{3x^{1/3}} \]
Combining the terms:
\[ f'(x) = 3x^{2/3} + \frac{2(3x - 7)}{3} \cdot \frac{1}{x^{1/3}} = 3x^{2/3} + \frac{2(3x - 7)}{3x^{1/3}} \]
Finding a common denominator:
\[ f'(x) = \frac{9x^{2/3}x^{1/3} + 2(3x - 7)}{3x^{1/3}} = \frac{9x + 6x - 14}{3x^{1/3}} = \frac{15x - 14}{3x^{1/3}} \]

### Step 3: Set the derivative to zero to find critical points

Setting \( f'(x) = 0 \):
\[ \frac{15x - 14}{3x^{1/3}} = 0 \]
This gives:
\[ 15x - 14 = 0 \implies x = \frac{14}{15} \]

### Step 4: Identify critical points and test intervals

The critical points are \( x = 0 \) (where the derivative is undefined) and \( x = \frac{14}{15} \). We will test the intervals:
1. \( (-\infty, 0) \)
2. \( (0, \frac{14}{15}) \)
3. \( (\frac{14}{15}, \infty) \)

### Step 5: Test the intervals

1. **For \( x < 0 \)** (e.g., \( x = -1 \)):
\[ f'(-1) = \frac{15(-1) - 14}{3(-1)^{1/3}} = \frac{-15 - 14}{3(-1)} = \frac{-29}{-3} > 0 \quad \text{(increasing)} \]

2. **For \( 0 < x < \frac{14}{15} \)** (e.g., \( x = 1 \)):
\[ f'(1) = \frac{15(1) - 14}{3(1)^{1/3}} = \frac{15 - 14}{3} = \frac{1}{3} > 0 \quad \text{(increasing)} \]

3. **For \( x > \frac{14}{15} \)** (e.g., \( x = 2 \)):
\[ f'(2) = \frac{15(2) - 14}{3(2)^{1/3}} = \frac{30 - 14}{3(2^{1/3})} = \frac{16}{3(2^{1/3})} > 0 \quad \text{(increasing)} \]

### Conclusion

The function \( f(x) \) is increasing in the intervals:
\[ (-\infty, 0) \cup \left(\frac{14}{15}, \infty\right) \]