Im having problem to find the second derivative , inflection point, concave up and down intervals.?

Given: # f(x) = (x-1)/(2x+1) = (N(x))/(D(x)) = (a_nx^n + ...)/(b_mx^m + ...)#

Analyze the graph analytically:

Find vertical asymptotes #D(x) = 0#:

#2x + 1 = 0; => x = -1/2#

Vertical asymptote at #x = -1/2#

Find horizontal asymptote:

#n = m => " horizontal asymptote ": y = a^n/b^n: " "y = 1/2#

Find the first derivative.
Use the quotient rule #(u/v)' = (vu' - uv')/v^2#

Let #u = x-1; " "u' = 1; " "v = 2x+1; " "v' = 2#

#f'(x) = ((2x+1)(1) - (x-1)(2))/(2x+1)^2#

#f'(x) = (2x + 1 -2x +2)/(2x+1)^2 = 3/(2x+1)^2#

Find the critical values #f'(x) = 0#:

#3/(2x+1)^2 = 0#

#3= 0" "# false

There is no relative maximum or minimum.

Find the second derivative using the power rule:

#f'(x) = 3/(2x+1)^2 = 3(2x+1)^-2#

#f''(x) = -6(2x+1)^-3 (2)#

#f''(x) = -12/(2x+1)^3#

Find inflection points by #f''(x) = 0#

#f''(x) = -12/(2x+1)^3 = 0#

#-12 = 0 " "# false

No inflection points

Find concavity:

This means concavity is only determined by the vertical asymptote

intervals: #(-oo, -1/2), (-1/2, oo)#

test value: #x = -1; " x = 0#

#f''(x) " "+ " " -#

concave up: #" "(-oo, -1/2)#

concave down: #" "(-1/2, oo)#