Ich gehe einmal ( ohne Überprüfung ) aus von
f(x) = ax^4 + bx^2 + c
f ´( x ) = 4ax^3 + 2bx
f(0) = 2
Einsetzen
f ( x ) = ax^4 + bx^2 + c
f ( 0 ) = a*(0)x^4 + b*(0)^2 + c = 2 => c = 2
f ( x ) = ax^4 + bx^2 + 2
f ´( x ) = 4ax^3 + 2bx
f ( 1 ) = a*1^4 + b*1^2 + 2 = 0
a + b + 2 = 0
f '(1) = 4*a*1^3 + 2*b*1 = 0
4a + 2b = 0
a + b + 2 = 0
a = -2 - b
4a = - 8 -4b
4a + 2b = 0
- 8 - 4b + 2b = 0
-2b - 8 = 0
b = -4
a + b + 2 = 0
a + -4 + 2 = 0
a = 2
f ( x ) = 2 * x^4 - 4*x^2 + 2