F(x)=5x^3/2-3x^5/2 is strictly increasing on interval

Dear student,

To check the interval in which given function is increasing we have differentiate the function first.

$$\begin{gathered} \mathrm{We} \mathrm{have}, \\ \mathrm{f}\left(\mathrm{x}\right) = \frac{5{\mathrm{x}}^3-3{\mathrm{x}}^5}{2} \\ \Rightarrow \mathrm{f}\text{'}\left(\mathrm{x}\right) = \frac{15{\mathrm{x}}^2-15{\mathrm{x}}^4}{2} \\ \mathrm{Now}, \mathrm{f}\text{'}\left(\mathrm{x}\right) = 0 \\ \Rightarrow \frac{15{\mathrm{x}}^2-15{\mathrm{x}}^4}{2} = 0 \\ \Rightarrow 15{\mathrm{x}}^2\left(1-{\mathrm{x}}^2\right) = 0 \\ \Rightarrow {\mathrm{x}}^2\left(1+\mathrm{x}\right)\left(1-\mathrm{x}\right) = 0 \\ \Rightarrow \mathrm{x} = 0, -1 \mathrm{and} 1 \\ \mathrm{The} \mathrm{points} 0, -1 \mathrm{and} 1 \mathrm{divide} \mathrm{the} \mathrm{real} \mathrm{line} \mathrm{into} 4 \mathrm{disjoint} \mathrm{intervals} \mathrm{namely} : \\ \left(-\infty , -1\right); \left(-1, 0\right); \left(0, 1\right); \left(1, \infty \right). \end{gathered}$$

INTERVALS	sign of x2	sign of (1+x)	sign of (1-x)	sign of f'(x)
(-$\infty$, -1)	+	-	+	-
(-1,0)	+	+	+	+
(0, 1)	+	+	+	+
(1, $\infty$)	+	+	-	-

Hence, the given function is strictly increasing in the interval (-1, 0) and (0, 1).

Regards