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 need to follow these steps:

### Step 1: Differentiate the function
We start by finding the derivative of \( f(x) \).

\[ f'(x) = \frac{d}{dx}[(3x - 7)x^{2/3}] \]

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:

\[ f'(x) = (3)(x^{2/3}) + (3x - 7)\left(\frac{2}{3}x^{-1/3}\right) \]

### Step 2: Simplify the derivative
Now we simplify \( f'(x) \):

\[ f'(x) = 3x^{2/3} + \frac{2}{3}(3x - 7)x^{-1/3} \]

Combining the terms:

\[ f'(x) = 3x^{2/3} + \frac{2(3x - 7)}{3x^{1/3}} \]

To combine these, we can express \( 3x^{2/3} \) as \( \frac{9x^{2/3}}{3} \):

\[ f'(x) = \frac{9x^{2/3}}{3} + \frac{2(3x - 7)}{3x^{1/3}} = \frac{9x^{2/3} + 2(3x - 7)x^{-1/3}}{3} \]

### Step 3: Set the derivative to zero
To find the critical points, we set \( f'(x) = 0 \):

\[ 9x^{2/3} + 2(3x - 7)x^{-1/3} = 0 \]

Multiplying through by \( 3x^{1/3} \) to eliminate the fraction:

\[ 27x + 2(3x - 7) = 0 \]

Expanding and simplifying:

\[ 27x + 6x - 14 = 0 \implies 33x - 14 = 0 \implies x = \frac{14}{33} \]

### Step 4: Analyze the intervals
Next, we need to check the sign of \( f'(x) \) around the critical points \( x = 0 \) and \( x = \frac{14}{33} \).

1. **For \( x < 0 \)** (e.g., \( x = -1 \)):
\[ f'(-1) = 3(-1)^{2/3} + \frac{2(3(-1) - 7)}{3(-1)^{1/3}} > 0 \quad \text{(positive)} \]

2. **For \( 0 < x < \frac{14}{33} \)** (e.g., \( x = 1 \)):
\[ f'(1) = 3(1)^{2/3} + \frac{2(3(1) - 7)}{3(1)^{1/3}} < 0 \quad \text{(negative)} \]

3. **For \( x > \frac{14}{33} \)** (e.g., \( x = 1 \)):
\[ f'(2) = 3(2)^{2/3} + \frac{2(3(2) - 7)}{3(2)^{1/3}} > 0 \quad \text{(positive)} \]

### Conclusion
The function \( f(x) \) is increasing in the intervals:

\[ (-\infty, 0) \cup \left(\frac{14}{33}, \infty\right) \]