Question

沐沐

19 20求解答還有過程🙏

f(x)=3x-6x = 0.
3入(X-2)
19.三次函數f(x)=x²+ax²+bx+c圖形的反曲點為(1,2),且f'(0)=2,試求實數數對(a,b,c) 的值。
20. f(x)=x3+3x²-7,求:過點(1,f(1))的切線方程式(以ax+by+c=0表示)

Answer 1

綾

19.
f(x) = 1(x-1)^3 + p(x-1) - 2
= x^3 - 3*x^2 + (3 + p)x - 5

a = -3
b = 3 + p
c = 5

f’(x) = 3*x^2 - 6x + b
f’(0) = b = 2

20.
f’(x) = 3*x^2 + 6x
f(1) = -3
f’(1) = 9
切線方程式：y - f(1) = f’(1)*(x - 1)
=> y - (-3) = 9(x - 1)
=> 9x - y - 12 = 0