For what values of x is dy/dx zero and undefined?

#y = "x^2 - 3x + 1"/"x+2"#

1 Answer
Feb 22, 2018

#dy/dx# is zero for #x =-2 pm sqrt(11)#, and #dy/dx# is undefined for #x=-2#

Explanation:

Find the derivative:
#dy/dx = (d(x^2 - 3x + 1))/dx 1/(x+2) + (x^2 - 3x + 1)(d)/(dx)(1/(x + 2))#
#=(2x-3)/(x+2) - (x^2 - 3x + 1)1/(x + 2)^2#
#=((2x-3)(x+2) - (x^2 - 3x + 1))/(x + 2)^2#
#= (2x^2 - 3x + 4x -6 - x^2+3x-1)/(x + 2)^2#
#= (x^2 + 4x -7)/(x + 2)^2#
by the product rule and various simplifications.

Find zeros:
#dy/dx=0# if and only if #x^2 + 4x -7=0#.
The roots of this polynomial are
#x_{1,2} = (1/2)(-4 pm sqrt(4^2 - 4(-7))) = -2 pm sqrt(11)#,
so #dy/dx=0# for #x = -2 pm sqrt(11)#.

Find where #dy/dx# is undefined:
Since division by #0# is not allowed, #dy/dx# is undefined where #(x + 2)^2=0#, that is, where
#x=-2#.