Answer:
[tex]\displaystyle f(x) = \frac{x + 3}{x - 2}[/tex] has no relative extrema when the domain is [tex]\mathbb{R} \backslash \lbrace 2 \rbrace[/tex] (the set of all real numbers other than [tex]2[/tex].)
Step-by-step explanation:
Assume that the domain of [tex]\displaystyle f(x) = \frac{x + 3}{x - 2}[/tex] is [tex]\mathbb{R} \backslash \lbrace 2 \rbrace[/tex] (the set of all real numbers other than [tex]2[/tex].)
Let [tex]f^{\prime}(x)[/tex] and [tex]f^{\prime\prime}(x)[/tex] denote the first and second derivative of this function at [tex]x[/tex].
Since this domain is an open interval, [tex]x = a[/tex] is a relative extremum of this function if and only if [tex]f^{\prime}(a) = 0[/tex] and [tex]f^{\prime\prime}(a) \ne 0[/tex].
Hence, if it could be shown that [tex]f^{\prime}(x) \ne 0[/tex] for all [tex]x \in \mathbb{R} \backslash \lbrace 2 \rbrace[/tex], one could conclude that it is impossible for [tex]\displaystyle f(x) = \frac{x + 3}{x - 2}[/tex] to have any relative extrema over this domain- regardless of the value of [tex]f^{\prime\prime}(x)[/tex].
[tex]\displaystyle f(x) = \frac{x + 3}{x - 2} = (x + 3) \, (x - 2)^{-1}[/tex].
Apply the product rule and the power rule to find [tex]f^{\prime}(x)[/tex].
[tex]\begin{aligned}f^{\prime}(x) &= \frac{d}{dx} \left[ (x + 3) \, (x - 2)^{-1}\right] \\ &= \left(\frac{d}{dx}\, [(x + 3)]\right)\, (x - 2)^{-1} \\ &\quad\quad (x + 3)\, \left(\frac{d}{dx}\, [(x - 2)^{-1}]\right) \\ &= (x - 2)^{-1} \\ &\quad\quad+ (x + 3) \, \left[(-1)\, (x - 2)^{-2}\, \left(\frac{d}{dx}\, [(x - 2)]\right) \right] \\ &= \frac{1}{x - 2} + \frac{-(x+ 3)}{(x - 2)^{2}} \\ &= \frac{(x - 2) - (x + 3)}{(x - 2)^{2}} = \frac{-5}{(x - 2)^{2}}\end{aligned}[/tex].
In other words, [tex]\displaystyle f^{\prime}(x) = \frac{-5}{(x - 2)^{2}}[/tex] for all [tex]x \in \mathbb{R} \backslash \lbrace 2 \rbrace[/tex].
Since the numerator of this fraction is a non-zero constant, [tex]f^{\prime}(x) \ne 0[/tex] for all [tex]x \in \mathbb{R} \backslash \lbrace 2 \rbrace[/tex]. (To be precise, [tex]f^{\prime}(x) < 0[/tex] for all [tex]x \in \mathbb{R} \backslash \lbrace 2 \rbrace\![/tex].)
Hence, regardless of the value of [tex]f^{\prime\prime}(x)[/tex], the function [tex]f(x)[/tex] would have no relative extrema over the domain [tex]x \in \mathbb{R} \backslash \lbrace 2 \rbrace[/tex].
