Points of Inflection on a rational function

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The discussion focuses on finding the inflection points of the rational function f(x) = (x - 1)/(x^2 - 4) by using the second derivative test. The second derivative, f''(x), is computed as (2x^3 - 6x^2 + 24x - 8)/(x^2 - 4)^4, but the numerator is proving difficult to solve for its roots. Attempts to factor and simplify lead to the cubic equation 2x^3 - 6x^2 + 24x - 8 = 0, which does not yield simple rational roots. The discussion highlights the challenge of solving the cubic equation and the realization that none of the rational root candidates work. Ultimately, the thread emphasizes the complexity of determining concavity and inflection points for this function.
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Homework Statement



Find the inflection points and use second derivative test to determine where the function is concave up or down

Homework Equations



f(x) = (x - 1)/(x2 - 4)

The Attempt at a Solution



f'(x) = (-x2 + 2x - 4) / (x2 - 4)2

f''(x) = 2x3 - 6x2 +24x -8 / (x2-4)4

This is where I am stuck. I can't solve the numerator set to 0. You can factor out the 2 and that's about it. Cubic that I can't solve. I'm looking at:

x(x2 - 3x +12) = 4

and thinking that maybe I can use the quadratic equation to find when that quadratic = 0 but I don't think that will help me.
 
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You want to solve 2x^3- 6x^2+ 24x- 8= 0? The obvious first thing to do is to divide through by 2 to get x^3- 3x^2+ 12x- 4= 0. Now the only possible rational roots are \pm 1, \pm 2, and \pm 4. Trying those in turn, we see that none of them satisfy the equation so there is not any "simple" solution.
 
Question: A clock's minute hand has length 4 and its hour hand has length 3. What is the distance between the tips at the moment when it is increasing most rapidly?(Putnam Exam Question) Answer: Making assumption that both the hands moves at constant angular velocities, the answer is ## \sqrt{7} .## But don't you think this assumption is somewhat doubtful and wrong?

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