If f(x) = x^{3/2}(3x - 10),x geq 0,then in wh | vedclass

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(B) Given function is $f(x) = x^{3/2}(3x - 10) = 3x^{5/2} - 10x^{3/2}$.To find the interval where the function is increasing,we find the derivative $f

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$[2, \infty)$

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(B) Given function is $f(x) = x^{3/2}(3x - 10) = 3x^{5/2} - 10x^{3/2}$.To find the interval where the function is increasing,we find the derivative $f'(x)$:$f'(x) = \frac{d}{dx}(3x^{5/2} - 10x^{3/2}) = 3 \cdot \frac{5}{2} x^{3/2} - 10 \cdot \frac{3}{2} x^{1/2} = \frac{15}{2} x^{3/2} - 15 x^{1/2}$.Setting $f'(x) = 0$ for critical points:$\frac{15}{2} x^{1/2} (x - 2) = 0$.Since $x \geq 0$,the critical point is $x = 2$.For $x > 2$,$f'(x) > 0$,so the function is increasing on $[2, \infty)$.For $0 < x < 2$,$f'(x) < 0$,so the function is decreasing on $(0, 2)$.

If $f(x) = x^{3/2}(3x - 10)$,$x \geq 0$,then in which interval is $f(x)$ an increasing function?

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