$-6$ . Non-GDC $\newline$ Let $f(x)=\frac{2-3 x^{5}}{2 x^{3}}, x \in \mathbb{R}, x \neq 0$ . $\newline$ (a) The graph of $y=f(x)$ has a local maximum at A. Find the coordinates of A.

Full solution

-6

. Non-GDC

\newline

Let

f(x)=\frac{2-3 x^{5}}{2 x^{3}}, x \in \mathbb{R}, x \neq 0

\newline

(a) The graph of

y=f(x)

has a local maximum at A. Find the coordinates of A.

Find Derivative and Simplify: To find the local maximum, we need to find the derivative of $f(x)$ and set it equal to zero. $\newline$ $f'(x) = \frac{d}{dx} \left[\frac{(2 - 3x^5)}{(2x^3)}\right]$ $\newline$ Using the quotient rule: $\frac{(v'u - uv')}{v^2}$ $\newline$ Let $u = (2 - 3x^5)$ and $v = 2x^3$ $\newline$ $u' = \frac{d}{dx} (2 - 3x^5) = -15x^4$ $\newline$ $v' = \frac{d}{dx} (2x^3) = 6x^2$ $\newline$ $f'(x) = \frac{[(2x^3)(-15x^4) - (2 - 3x^5)(6x^2)]}{(2x^3)^2}$
Set Derivative Equal to Zero: Simplify the derivative. $\newline$ $f'(x) = \frac{-30x^7 - (12x^2 - 18x^7)}{4x^6}$ $\newline$ $f'(x) = \frac{-30x^7 - 12x^2 + 18x^7}{4x^6}$ $\newline$ $f'(x) = \frac{-12x^7 - 12x^2}{4x^6}$ $\newline$ $f'(x) = \frac{-3x^7}{4x^6} - \frac{3x^2}{4x^6}$ $\newline$ $f'(x) = -\frac{3x}{4} - \frac{3}{4x^4}$
Solve for Critical Points: Set the derivative equal to zero to find critical points. $\newline$ $-\frac{3x}{4} - \frac{3}{4}x^4 = 0$ $\newline$ Multiply through by $4x^4$ to clear the fraction. $\newline$ $-3x^5 - 3 = 0$