[Risolto] Derivate e problemi sulle funzioni

FORME PARAMETRICHE
* f(x) = y = (a*x^2 + b - 2)/(x^2 + c)
* f'(x) = 2*(a*c - b + 2)*x/(x^2 + c)^2
* f''(x) = - 2*(3*x^2 - c)*(a*c - b + 2)/(x^2 + c)^3
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CONDIZIONI
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A) flesso in F(2, - 3) ≡ (f(2) = - 3) & (f''(2) = 0)
* f''(2) = - 2*(3*2^2 - c)*(a*c - b + 2)/(2^2 + c)^3 = 0
* f(2) = (a*2^2 + b - 2)/(2^2 + c) = - 3
* (- 2*(3*2^2 - c)*(a*c - b + 2)/(2^2 + c)^3 = 0) & ((a*2^2 + b - 2)/(2^2 + c) = - 3) ≡
≡ (b = - 2*(2*a + 23)) & (c = 12)
da cui
* f(x) = y = (a*x^2 - 4*(a + 12))/(x^2 + 12)
* f'(x) = 32*(a + 3)*x/(x^2 + 12)^2
* f''(x) = - 96*(a + 3)*(x^2 - 4)/(x^2 + 12)^3
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B) estremo relativo in y = - 1 ≡ (f'(x) = 0) & (f''(x) != 0)
* f(x) = y = (a*x^2 - 4*(a + 12))/(x^2 + 12) = - 1 ≡
≡ ((a + 1)*x^2 - 4*(a + 9))/(x^2 + 12) = 0 ≡
≡ x^2 = 4*(a + 9)/(a + 1) ≡
≡ x = ± 2*√((a + 9)/(a + 1))
quindi
* (f'(± 2*√((a + 9)/(a + 1))) = 0) & (f''(± 2*√((a + 9)/(a + 1))) != 0) ≡
≡ ((a + 3)*(± 2*√((a + 9)/(a + 1))) = 0) & ((a + 3)/(a + 1) != 0) ≡
≡ ((a + 9)/(a + 1) = 0) & ((a + 3)/(a + 1) != 0) ≡
≡ a = - 9
da cui
* b = - 10
* f(x) = y = - 3*(3*x^2 + 4)/(x^2 + 12)
* f'(x) = - 192*x/(x^2 + 12)^2
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CONCLUSIONE: se i calcoli sono Ok allora le tangenti devono tangere il giusto.
* f'(2) = - 3/2
* tF ≡ y = - 3 - (3/2)*(x - 2) ≡ y = - (3/2)*x
* tE ≡ y = - 1
http://www.wolframalpha.com/input?i=%5B%28-%283%2F2%29*x-y%29*%28-1-y%29%3D0%2Cy%3D-3*%283*x%5E2--4%29%2F%28x%5E2--12%29%5D
http://www.wolframalpha.com/input?i=%5B%28-%283%2F2%29*x-y%29*%28-1-y%29%3D0%2Cy%3D-3*%283*x%5E2--4%29%2F%28x%5E2--12%29%5Dx%3D-19to19%2Cy%3D-11to0