# Thread: determining the maximum and minimum

1. ## determining the maximum and minimum

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

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

\displaystyle \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 $\displaystyle x=2$, and as the function decreases after $\displaystyle x=2$, a maximum occurs here.

3. 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.

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

5. 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 ... $\displaystyle y' = \frac{2-x}{x^3} < 0$ for all $\displaystyle x < 0$