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Given the function $f(x) = 5x^{\frac{3}{2}} - 3x^{\frac{5}{2}}$, we need to find its derivative $f'(x)$. $$f'(x) = \frac{d}{dx}(5x^{\frac{3}{2}} - 3x^{\frac{5}{2}})$$ $$f'(x) = 5 \cdot \frac{3}{2}x^{\frac{3}{2}-1} - 3 \cdot \frac{5}{2}x^{\frac{5}{2}-1}$$ $$f'(x) = \frac{15}{2}x^{\frac{1}{2}} - \frac{15}{2}x^{\frac{3}{2}}$$ $$f'(x) = \frac{15}{2}x^{\frac{1}{2}}(1 - x)$$
To find the intervals where the function is increasing or decreasing, we need to find the critical points by setting $f'(x) = 0$. $$\frac{15}{2}x^{\frac{1}{2}}(1 - x) = 0$$ This gives us $x^{\frac{1}{2}} = 0$ or $1 - x = 0$. So, $x = 0$ or $x = 1$. Also, we need to consider the domain of the function. Since we have $x^{\frac{3}{2}}$ and $x^{\frac{5}{2}}$, $x$ must be non-negative. Thus, $x \ge 0$.
We have the critical points $x = 0$ and $x = 1$. These points divide the domain $x \ge 0$ into the intervals $(0, 1)$ and $(1, \infty)$.
We will test the sign of $f'(x)$ in each interval. Interval $(0, 1)$: Choose a test point, say $x = 0.5$. $$f'(0.5) = \frac{15}{2}(0.5)^{\frac{1}{2}}(1 - 0.5) = \frac{15}{2}\sqrt{0.5}(0.5) > 0$$ Since $f'(x) > 0$ in $(0, 1)$, the function is increasing in this interval. Interval $(1, \infty)$: Choose a test point, say $x = 4$. $$f'(4) = \frac{15}{2}(4)^{\frac{1}{2}}(1 - 4) = \frac{15}{2}(2)(-3) = -45 < 0$$ Since $f'(x) < 0$ in $(1, \infty)$, the function is decreasing in this interval.
(i) Increasing: $(0, 1)$ (ii) Decreasing: $(1, \infty)$
Final Answer: Increasing: (0, 1), Decreasing: (1, ∞)
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