Key Concepts and Formulas

• A function $f(x)$ is increasing when its first derivative $f'(x) > 0$.
• Product Rule: If $f(x) = u(x)v(x)$, then $f'(x) = u'(x)v(x) + u(x)v'(x)$.
• Power Rule: $\frac{d}{dx}(x^n) = nx^{n-1}$.

Step-by-Step Solution

1. Identify the Given Function and Goal

We are given the function $f(x) = (3x - 7)x^{2/3}$. Our goal is to find the intervals where $f(x)$ is increasing, meaning we need to find where $f'(x) > 0$.

2. Differentiate the Function Using the Product Rule

The function $f(x)$ is a product of two functions: $u(x) = 3x - 7$ and $v(x) = x^{2/3}$. We will apply the product rule: $f'(x) = u'(x)v(x) + u(x)v'(x)$.

• Find $u'(x)$: $u(x) = 3x - 7$ $u'(x) = \frac{d}{dx}(3x - 7) = 3$

• Find $v'(x)$: $v(x) = x^{2/3}$ $v'(x) = \frac{d}{dx}(x^{2/3}) = \frac{2}{3}x^{(2/3 - 1)} = \frac{2}{3}x^{-1/3} = \frac{2}{3x^{1/3}}$

• Apply the Product Rule: $f'(x) = u'(x)v(x) + u(x)v'(x) = (3)(x^{2/3}) + (3x - 7)\left(\frac{2}{3x^{1/3}}\right)$ $f'(x) = 3x^{2/3} + \frac{2(3x - 7)}{3x^{1/3}}$

3. Simplify the Expression for $f'(x)$

To simplify, we find a common denominator, which is $3x^{1/3}$: $f'(x) = \frac{3x^{2/3}(3x^{1/3})}{3x^{1/3}} + \frac{2(3x - 7)}{3x^{1/3}}$ $f'(x) = \frac{9x^{2/3 + 1/3} + 6x - 14}{3x^{1/3}}$ Since $x^{2/3} \cdot x^{1/3} = x$, $f'(x) = \frac{9x + 6x - 14}{3x^{1/3}} = \frac{15x - 14}{3x^{1/3}}$

4. Set up the Inequality for an Increasing Function

For $f(x)$ to be increasing, we need $f'(x) > 0$: $\frac{15x - 14}{3x^{1/3}} > 0$

5. Solve the Inequality

Find the critical points by setting the numerator and denominator equal to zero:

• Numerator: $15x - 14 = 0 \Rightarrow x = \frac{14}{15}$
• Denominator: $3x^{1/3} = 0 \Rightarrow x = 0$

These critical points, $x = 0$ and $x = \frac{14}{15}$, divide the real number line into three intervals: $(-\infty, 0)$, $(0, \frac{14}{15})$, and $(\frac{14}{15}, \infty)$. We test a value from each interval in the inequality $\frac{15x - 14}{3x^{1/3}} > 0$:

• Interval $(-\infty, 0)$: Let $x = -1$. Then $\frac{15(-1) - 14}{3(-1)^{1/3}} = \frac{-29}{-3} = \frac{29}{3} > 0$. So, $f'(x) > 0$ on $(-\infty, 0)$.
• Interval $(0, \frac{14}{15})$: Let $x = \frac{1}{2}$. Then $\frac{15(\frac{1}{2}) - 14}{3(\frac{1}{2})^{1/3}} = \frac{7.5 - 14}{3(\frac{1}{2})^{1/3}} = \frac{-6.5}{3(\frac{1}{2})^{1/3}} < 0$. So, $f'(x) < 0$ on $(0, \frac{14}{15})$.
• Interval $(\frac{14}{15}, \infty)$: Let $x = 1$. Then $\frac{15(1) - 14}{3(1)^{1/3}} = \frac{1}{3} > 0$. So, $f'(x) > 0$ on $(\frac{14}{15}, \infty)$.

Therefore, $f(x)$ is increasing on the intervals $(-\infty, 0)$ and $(\frac{14}{15}, \infty)$.

Common Mistakes & Tips

• Remember to consider where the denominator of $f'(x)$ is zero, as this can also define intervals where the function's behavior changes.
• Be careful with signs when testing intervals for inequalities. A simple sign error can lead to the wrong answer.
• Don't forget the product rule when differentiating a product of functions.

Summary

We found the intervals where the given function $f(x) = (3x - 7)x^{2/3}$ is increasing by finding its derivative $f'(x)$, setting up the inequality $f'(x) > 0$, and solving for $x$. We identified critical points at $x=0$ and $x=\frac{14}{15}$, tested the intervals defined by these points, and found that $f(x)$ is increasing on $(-\infty, 0) \cup (\frac{14}{15}, \infty)$.

Final Answer

The final answer is $\boxed{\left( { - \infty ,0} \right) \cup \left( {{{14} \over {15}},\infty } \right)}$, which corresponds to option (A).