Answer
$f'(x)= \frac{x\cos(x)-\sin(x)}{x^{2}}$
$g'(x)= \frac{x\cos(x)-\sin(x)}{x^{2}}$
The slope of $f(x)$ is equal to the slope of $g(x)$. They are parallel functions.
Work Step by Step
Quotient rule: $\frac{d}{dx}$$\frac{f(x)}{g(x)}$=$\frac{g(x)f'(x)-f(x)g'(x)}{g(x)g(x)}$
$f(x)= \frac{sin(x)-3x}{x}$
$f'(x)= \frac{(x)[cos(x)-3]-[sin(x)-3x](1)}{x^{2}}$=$\frac{xcos(x)-3x-sin(x)+3x}{x^{2}}$=$\frac{xcos(x)-sin(x)}{x^{2}}$
$g(x)=\frac{sin(x)+2x}{x}$
$g'(x)=\frac{(x)[cos(x)+2]-[sin(x)-2x](1)}{x^{2}}$=$\frac{cos(x)+2x-sin(x)-2x}{x^{2}}$=$\frac{xcos(x)-sin(x)}{x^{2}}$
The slope of $f(x)$ is equal to the slope of $g(x)$. They are parallel functions.
