求a,b,c的值 ,使f(x)=ax^3+bx^2+cx有一拐点(1,2)且在该点的切线斜率为-1
问题描述：
求a,b,c的值 ,使f(x)=ax^3+bx^2+cx有一拐点(1,2)且在该点的切线斜率为-1
切线斜率的条件怎么用
答
f(x) = ax³ + bx² + cx,f(1) = a + b + c = 2 (1)f'(x) = 3ax² + 2bx + c,f'(1) = 3a + 2b + c = -1 (2) (切线斜率)f"(x) = 6ax + 2b,f"(1) = 6a + 2b = 0 (拐点) (3)由(1)-(3):a = 3,b = -9,c = 8...