determining the maximum and minimum

**wisezeta** · December 21st 2009, 03:32 PM

does the function y= (x-1)/x^2 have a maximum or a minimum because when i conduct a derivative test there signs change from positive to negative at x=2, but when i graph the function on a graphing calculator there doesn't seem to be a maximum

**Scott H** · December 21st 2009, 03:52 PM

The function does have a maximum, and we may calculate the derivative as follows:

$\begin{aligned} \frac{d}{dx}\frac{x-1}{x^2}&=\frac{d}{dx}(x^{-1}-x^{-2})\\ &=\frac{d}{dx}(x^{-1})-\frac{d}{dx}(x^{-2})\\ &=-x^{-2}+2x^{-3}. \end{aligned} $

You are correct that the derivative changes sign at $x=2$ , and as the function decreases after $x=2$ , a maximum occurs here.

**skeeter** · December 21st 2009, 03:53 PM

Originally Posted by wisezeta

does the function y= (x-1)/x^2 have a maximum or a minimum because when i conduct a derivative test there signs change from positive to negative at x=2, but when i graph the function on a graphing calculator there doesn't seem to be a maximum

recheck your graph in closer window ... the max is subtle.

**wisezeta** · December 21st 2009, 04:00 PM

Ok, but what about the 3rd quadrant. Would there be an extrema there as well? It is similar to the extrema at x=2

**skeeter** · December 21st 2009, 04:04 PM

Originally Posted by wisezeta

Ok, but what about the 3rd quadrant. Would there be an extrema there as well? It is similar to the extrema at x=2

no extrema in quad III ... $y' = \frac{2-x}{x^3} < 0$ for all $x < 0$

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