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

Question

Scarlett Allison · Accepted Answer

#y# has a local minimum of 4 at #x=1/2#

#y=1/x - 1/(x-1)#

#= x^(-1) - (x-1)^(-1)#

#y' = -x^-2 + (x-1)^-2 *1# {Power rule and Chain rule]

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

I.e. where: #-x^-2 + (x-1)^-2 =0#

#(-(x-1)^2+x^2)/(x^2(x-1)^2)=0#

#-x^2+2x-1 + x^2 =0#

#x=1/2#

As can be seen from the graph of #y# below, #x=1/2# is local minimum.

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

Thus: #y_min = y(1/2) = 1/(1/2) - 1/(1/2 -1)#

#= 2+2 =4#

Scarlett Allison · Answer

undefined To find the local extrema for $ y = \frac{1}{x} - \frac{1}{x - 1} $, follow these steps:

1. Find the first derivative of $ y $ with respect to $ x $.
2. Set the derivative equal to zero and solve for $ x $.
3. Check the second derivative at the critical points found in step 2.
4. Determine the nature of the local extrema based on the signs of the second derivative.

Let's proceed with these steps:

1. Find the first derivative $ y' $:
$$ y' = -\frac{1}{x^2} + \frac{1}{(x - 1)^2} $$

2. Set $ y' = 0 $ and solve for $ x $:
$$ -\frac{1}{x^2} + \frac{1}{(x - 1)^2} = 0 $$

3. Solve the equation to find critical points $ x $.
   $$ \frac{1}{(x - 1)^2} = \frac{1}{x^2} $$
   $$ (x - 1)^2 = x^2 $$
   $$ x^2 - 2x + 1 = x^2 $$
   $$ -2x + 1 = 0 $$
   $$ x = \frac{1}{2} $$

4. Check the second derivative at $ x = \frac{1}{2} $:
   $$ y'' = \frac{2}{x^3} - \frac{2}{(x - 1)^3} $$
   Plug in $ x = \frac{1}{2} $:
   $$ y'' = \frac{2}{(\frac{1}{2})^3} - \frac{2}{(\frac{1}{2} - 1)^3} = 16 > 0 $$

Since the second derivative is positive at $ x = \frac{1}{2} $, the function has a local minimum at $ x = \frac{1}{2} $.