21/04/18 02:48:48.05 T2x+9b6/.net
t を a,bで表わせ、なんて問題はまさか出ないだろうけど…
g(0), g(a) が極値だから x=a/2 が変曲点。
g(t) = 2t^3 - 3a･t^2 + g(0)
= 2(t - a/2)^3 - 6(a/2)^2･(t - a/2) + {g(0)+g(a)}/2,
∴ t = {a + (√[-g(0)] + √[-g(a)])^(2/3) + (√[-g(0)] - √[-g(a)])^(2/3)} /2,
21/04/18 02:48:48.05 T2x+9b6/.net
t を a,bで表わせ、なんて問題はまさか出ないだろうけど…
g(0), g(a) が極値だから x=a/2 が変曲点。
g(t) = 2t^3 - 3a･t^2 + g(0)
= 2(t - a/2)^3 - 6(a/2)^2･(t - a/2) + {g(0)+g(a)}/2,
∴ t = {a + (√[-g(0)] + √[-g(a)])^(2/3) + (√[-g(0)] - √[-g(a)])^(2/3)} /2,