Hallo Peter,
f(x) = - x3 + 4·x2 - 3·x (Graph: links oben beginnende S-Form)
g(x) = a * (x-1) * (x-2) , S(2|-0.2) einsetzen → a = - 0,5
g(x) = - 0,5 · (x - 1) * (x - 2) = 1/2 · x2 - 2·x + 3/2
f '(x) = - 3·x2 + 8·x - 3 = 0 →abc-Formel xH ≈ 2,215 ; [ xT ≈ 0,451 ]
f(2,215) = 2.113 Hf = ( 2,215 | 2.113)
Tg = S = ( 2 | - 0,5 )
Pythagoras →
gesuchte Seelänge = √( Δx2 + Δy2 ) = √( 0,2152 + 2,6132 )
≈ 2.622 [km] ≈ 2622 m
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Fläche = 1∫3 ( - x3 + 4·x2 - 3·x - ( 1/2 · x2 - 2·x + 3/2 ) ) dx
= ... = [ - x4/4 + 7·x3/6 - x2/2 - 3·x/2 ]13 = 10/3
Gruß Wolfgang