1. Given function

We need the interval where $f(x)=(3x-7)x^{2/3}, \, x\in \mathbb R$ is increasing.

2. Differentiate the function

Using product rule: $f(x)=(3x-7)x^{2/3}$

So, $f'(x)=3x^{2/3}+(3x-7)\cdot \frac{2}{3}x^{-1/3}$

Take common factor $x^{-1/3}$: $f'(x)=x^{-1/3}\left(3x+\frac{2}{3}(3x-7)\right)$

Simplify the bracket: $3x+\frac{2}{3}(3x-7)=3x+2x-\frac{14}{3}=5x-\frac{14}{3}$

Hence, $f'(x)=x^{-1/3}\left(5x-\frac{14}{3}\right)$

Or, $f'(x)=\frac{1}{3}x^{-1/3}(15x-14)$

3. Critical points

For monotonicity, check where $f'(x)=0$ or undefined.

• $f'(x)=0$ when $15x-14=0 \implies x=\frac{14}{15}$
• $f'(x)$ is undefined at $x=0$

So divide the real line into intervals: $(-\infty,0), \quad (0,\tfrac{14}{15}), \quad (\tfrac{14}{15},\infty)$

4. Sign analysis of $f'(x)$

Since $f'(x)=\frac{1}{3}x^{-1/3}(15x-14),$ the sign depends on $x^{-1/3}$ and $(15x-14)$.

• For $x<0$:

  • $x^{-1/3}<0$
  • $15x-14<0$
  • Product is positive

  Therefore, $f'(x)>0$ on $(-\infty,0)$.

• For $0<x<\frac{14}{15}$:

  • $x^{-1/3}>0$
  • $15x-14<0$
  • Product is negative

  Therefore, $f'(x)<0$ on $(0,\tfrac{14}{15})$.

• For $x>\frac{14}{15}$:

  • $x^{-1/3}>0$
  • $15x-14>0$
  • Product is positive

  Therefore, $f'(x)>0$ on $(\tfrac{14}{15},\infty)$.

5. Conclusion

The function is increasing where $f'(x)>0$, i.e. $(-\infty,0)\cup\left(\frac{14}{15},\infty\right)$

6. Match with options

This is Option B.

7. Comparison with stored answer

Stored correct answer: B

Our derived answer: B

So they agree.