How to Find the Inflection Points of this Function

Find when the 2nd derivative equals 0

f(x) = (6x^2 + 5x - 4) / (2x^2 - 7x - 4)

u = 6x^2 + 5x - 4
du = 12x + 5
v = 2x^2 - 7x - 4
dv = 4x - 7

f'(x) =>
(v * du - u * dv) / v^2 =>
((2x^2 - 7x - 4) * (12x + 5) - (6x^2 + 5x - 4) * (4x - 7)) / (2x^2 - 7x - 4)^2 =>
(24x^3 + 10x^2 - 84x^2 - 35x - 48x - 20 - (24x^3 - 42x^2 + 20x^2 - 35x - 16x + 28) / (2x^2 - 7x - 4)^2 =>
(24x^3 - 24x^3 - 74x^2 + 22x^2 - 83x + 51x - 20 - 28) / (2x^2 - 7x - 4)^2 =>
(-52x^2 - 32x - 48) / (2x^2 - 7x - 4)^2 =>
(-4) * (13x^2 + 8x + 12) / (2x^2 - 7x - 4)^2

u = 13x^2 + 8x + 12
du = 26x + 8
v = (2x^2 - 7x - 4)^2
dv = 2 * (2x^2 - 7x - 4) * (4x - 7)

f''(x) =>
-4 * ((2x^2 - 7x - 4)^2 * (26x + 8) - (13x^2 + 8x + 12) * 2 * (4x - 7) * (2x^2 - 7x - 4)) / (2x^2 - 7x - 4)^4 =>
-8 * ((2x^2 - 7x - 4)^2 * (13x + 4) - (13x^2 + 8x + 12) * (4x - 7) * (2x^2 - 7x - 4)) / (2x^2 - 7x - 4)^4 =>
-8 * ((2x^2 - 7x - 4) * (13x + 4) - (13x^2 + 8x + 12) * (4x - 7)) / (2x^2 - 7x - 4)^3

Set the numerator to equal 0

(2x^2 - 7x - 4) * (13x + 4) - (13x^2 + 8x + 12) * (4x - 7) = 0
26x^3 + 8x^2 - 91x^2 - 28x - 52x - 16 - (52x^3 - 91x^2 + 32x^2 - 56x + 48x - 84) = 0
26x^3 - 52x^3 + 8x^2 - 91x^2 + 91x^2 - 32x^2 - 80x + 8x - 16 + 84 = 0
-26x^3 - 24x^2 - 72x + 68 = 0
13x^3 + 12x^2 + 36x - 34 = 0

http://www.wolframalpha.com/input/?i=13x…

A necessary condition for an inflection point is that the second derivative must equal zero. To find the values of x where the second derivative is zero, you must:

1. compute the first derivative.
2. compute the second derivative by taking the derivative of the first derivative.