Question

Relative extrema of f(x)=(x+3)/(x-2)

Answer 1

`\displaystyle f(x) = (x + 3)/(x - 2)` has no relative extrema when the domain is
`\mathbb{R} \backslash \lbrace 2 \rbrace` (the set of all real numbers other than
`2`.)

Explanation:

Assume that the domain of
`\displaystyle f(x) = (x + 3)/(x - 2)` is
`\mathbb{R} \backslash \lbrace 2 \rbrace` (the set of all real numbers other than
`2`.)

Let
`f^(\prime)(x)` and
`f^(\prime\prime)(x)` denote the first and second derivative of this function at
`x`.

Since this domain is an open interval,
`x = a` is a relative extremum of this function if and only if
`f^(\prime)(a) = 0` and
`f^(\prime\prime)(a) \\e 0`.

Hence, if it could be shown that
`f^(\prime)(x) \\e 0` for all
`x \in \mathbb{R} \backslash \lbrace 2 \rbrace`, one could conclude that it is impossible for
`\displaystyle f(x) = (x + 3)/(x - 2)` to have any relative extrema over this domain- regardless of the value of
`f^(\prime\prime)(x)`.

`\displaystyle f(x) = (x + 3)/(x - 2) = (x + 3) \, (x - 2)^(-1)`.

Apply the product rule and the power rule to find
`f^(\prime)(x)`.

`\begin{aligned}f^(\prime)(x) &= (d)/(dx) \left[ (x + 3) \, (x - 2)^(-1)\right] \\ &= \left((d)/(dx)\, [(x + 3)]\right)\, (x - 2)^(-1) \\ &\quad\quad (x + 3)\, \left((d)/(dx)\, [(x - 2)^(-1)]\right) \\ &= (x - 2)^(-1) \\ &\quad\quad+ (x + 3) \, \left[(-1)\, (x - 2)^(-2)\, \left((d)/(dx)\, [(x - 2)]\right) \right] \\ &= (1)/(x - 2) + (-(x+ 3))/((x - 2)^(2)) \\ &= ((x - 2) - (x + 3))/((x - 2)^(2)) = (-5)/((x - 2)^(2))\end{aligned}`.

In other words,
`\displaystyle f^(\prime)(x) = (-5)/((x - 2)^(2))` for all
`x \in \mathbb{R} \backslash \lbrace 2 \rbrace`.

Since the numerator of this fraction is a non-zero constant,
`f^(\prime)(x) \\e 0` for all
`x \in \mathbb{R} \backslash \lbrace 2 \rbrace`. (To be precise,
`f^(\prime)(x) < 0` for all
`x \in \mathbb{R} \backslash \lbrace 2 \rbrace\!`.)

Hence, regardless of the value of
`f^(\prime\prime)(x)`, the function
`f(x)` would have no relative extrema over the domain
`x \in \mathbb{R} \backslash \lbrace 2 \rbrace`.