Question:
The function, \(f(x) = (3x - 7)x^{2/3}\), \(x \in \mathbb{R}\), is increasing for all \(x\) lying in :
The function, \(f(x) = (3x - 7)x^{2/3}\), \(x \in \mathbb{R}\), is increasing for all \(x\) lying in :
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Write \(f(x) = 3x^{5/3} - 7x^{2/3}\), find \(f'(x)\) and determine where \(f'(x) \gt 0\) (note \(f'\) is undefined at 0).
Updated On: Nov 5, 2026
- \((-\infty, 0) \cup \left(\frac{3}{7}, \infty\right)\)
- \((-\infty, 0) \cup \left(\frac{14}{15}, \infty\right)\)
- \(\left(-\infty, \frac{14}{15}\right)\)
- \(\left(-\infty, \frac{14}{15}\right) \cup (0, \infty)\)
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The Correct Option is B
Solution and Explanation
\(f(x) = 3x^{5/3} - 7x^{2/3}\)
\(f'(x) = 5x^{2/3} - \frac{14}{3}x^{-1/3} = \frac{15x - 14}{3x^{1/3}}\)
f is increasing where \(f'(x) \gt 0\).
Critical points: \(x = 0\) (f' undefined) and \(x = \frac{14}{15}\).
For \(x \lt 0\): \(15x - 14 \lt 0\) and \(x^{1/3} \lt 0\), so \(f' \gt 0\).
For \(0 \lt x \lt \frac{14}{15}\): numerator negative, denominator positive, so \(f' \lt 0\).
For \(x \gt \frac{14}{15}\): both positive, so \(f' \gt 0\).
Hence f is increasing on \((-\infty, 0) \cup \left(\frac{14}{15}, \infty\right)\).
Answer: (2) \((-\infty, 0) \cup \left(\frac{14}{15}, \infty\right)\)
\(f'(x) = 5x^{2/3} - \frac{14}{3}x^{-1/3} = \frac{15x - 14}{3x^{1/3}}\)
f is increasing where \(f'(x) \gt 0\).
Critical points: \(x = 0\) (f' undefined) and \(x = \frac{14}{15}\).
For \(x \lt 0\): \(15x - 14 \lt 0\) and \(x^{1/3} \lt 0\), so \(f' \gt 0\).
For \(0 \lt x \lt \frac{14}{15}\): numerator negative, denominator positive, so \(f' \lt 0\).
For \(x \gt \frac{14}{15}\): both positive, so \(f' \gt 0\).
Hence f is increasing on \((-\infty, 0) \cup \left(\frac{14}{15}, \infty\right)\).
Answer: (2) \((-\infty, 0) \cup \left(\frac{14}{15}, \infty\right)\)
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