How do you find the local extrema for y = [1 / x] - [1 / (x - 1)]?

Mar 8, 2017

$y$ has a local minimum of 4 at $x = \frac{1}{2}$

Explanation:

$y = \frac{1}{x} - \frac{1}{x - 1}$

$= {x}^{- 1} - {\left(x - 1\right)}^{- 1}$

$y ' = - {x}^{-} 2 + {\left(x - 1\right)}^{-} 2 \cdot 1$ {Power rule and Chain rule]

$y$ will have extrema where: $y ' = 0$

I.e. where: $- {x}^{-} 2 + {\left(x - 1\right)}^{-} 2 = 0$

$\frac{- {\left(x - 1\right)}^{2} + {x}^{2}}{{x}^{2} {\left(x - 1\right)}^{2}} = 0$

$- {x}^{2} + 2 x - 1 + {x}^{2} = 0$

$x = \frac{1}{2}$

As can be seen from the graph of $y$ below, $x = \frac{1}{2}$ is local minimum.

graph{1/x - 1/(x-1) [-10.66, 11.84, -3.845, 7.405]}

Thus: ${y}_{\min} = y \left(\frac{1}{2}\right) = \frac{1}{\frac{1}{2}} - \frac{1}{\frac{1}{2} - 1}$

$= 2 + 2 = 4$