Given: #f(x) = x/(x^2-9) = x/((x-3)(x+3))#
Find critical values: #f'(x) = 0#:
Find the first derivative using the quotient rule:
# (u/v)' = (v u' - u v')/v^2#
Let #u = x; " "u' = 1#
Let #v = x^2-9; " "v' = 2x#
#f'(x) = ((x^2-9)(1) - x(2x))/(x^2-9)^2 = (x^2 - 2x^2 -9)/(x^2-9)^2 = (-x^2 - 9)/(x^2-9)^2 #
#f'(x) = (-(x^2 +9))/(x^2-9)^2 = 0#
#-(x^2 +9) = 0; " "x^2 + 9 = 0; " "=> x = +-3i#
No critical values.
Inflection points are found when #f''(x) = 0#
Find the second derivative:
Let #u = -x^2 - 9; " "u' = -2x#
Let #v = (x^2 - 9)^2; " "v' = 2(x^2-9)(2x) = 4x(x^2-9)#
Find #f''(x) = (((x^2 - 9)^2)(-2x) - (-x^2 - 9)(4x(x^2-9)))/((x^2-9)^4)#
#f''(x) = (((x^2 - 9)^2)(-2x) + (x^2 + 9)(4x(x^2-9)))/((x^2-9)^4)#
#f''(x) = ((x^2-9)[-2x(x^2-9) + 4x(x^2+9)])/((x^2-9)^4)#
#f''(x) = (-2x^3+18x + 4x^3 + 36x)/(x^2-9)^3 = (2x^3 + 54x)/(x^2-9)^3#
#f''(x) = (2x(x^2 + 27))/(x^2-9)^3 = 0#
#2x(x^2 + 27) = 0#
#x = 0, x = +- sqrt(27)i = +- 3sqrt(3)i#
Inflection point at #x = 0#
#f(0) = 0#
Inflection at #(0, 0)#