Answers: 2
has no relative extrema when the domain is 
(the set of all real numbers other than 
.)
and 
denote the first and second derivative of this function at 
.
is a relative extremum of this function if and only if 
and 
.
for all 
, one could conclude that it is impossible for 
.![\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}](/tpl/images/1393/9399/98966.png)
.
for all 
for all 
.)
would have no relative extrema over the domain 
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Relative extrema of f(x)=(x+3)/(x-2)...
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