The function,f(x)=(3x-7)x^{2/3}, x in R, is i | vedclass

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(B) Given function: $f(x) = (3x - 7)x^{2/3} = 3x^{5/3} - 7x^{2/3}$.To find the intervals where the function is increasing,we find the derivative $f'(x

Vedclass · Accepted Answer

$(-\infty, 0) \cup \left(\frac{14}{15}, \infty\right)$

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(B) Given function: $f(x) = (3x - 7)x^{2/3} = 3x^{5/3} - 7x^{2/3}$.To find the intervals where the function is increasing,we find the derivative $f'(x)$:$f'(x) = 3 \cdot \frac{5}{3}x^{2/3} - 7 \cdot \frac{2}{3}x^{-1/3}$$f'(x) = 5x^{2/3} - \frac{14}{3x^{1/3}}$Simplify the expression for $f'(x)$:$f'(x) = \frac{5x^{2/3} \cdot 3x^{1/3} - 14}{3x^{1/3}} = \frac{15x - 14}{3x^{1/3}}$For the function to be increasing,we require $f'(x) > 0$:$\frac{15x - 14}{3x^{1/3}} > 0$We analyze the sign of $f'(x)$ using the critical points $x = 0$ and $x = \frac{14}{15}$:- For $x  0$.- For $0  0$,so $f'(x) - For $x > \frac{14}{15}$,$15x - 14 > 0$ and $3x^{1/3} > 0$,so $f'(x) > 0$.Thus,$f(x)$ is increasing for $x \in (-\infty, 0) \cup \left(\frac{14}{15}, \infty\right)$.

The function,$f(x)=(3x-7)x^{2/3}, x \in R,$ is increasing for all $x$ lying in

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