已知三次函数f(x)=x3+ax2+bx+c在(-∞,-1).(2.+∞)上单调递增,在(-1,2)上单调递已知三次函数f(x)=x3+ax2+bx+c在(-∞,-1).(2.+∞)上单调递增,在(-1,2)上单调递减,若当且仅当x>4,f(x)>x2-4x+5,求f(x)的解
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![已知三次函数f(x)=x3+ax2+bx+c在(-∞,-1).(2.+∞)上单调递增,在(-1,2)上单调递已知三次函数f(x)=x3+ax2+bx+c在(-∞,-1).(2.+∞)上单调递增,在(-1,2)上单调递减,若当且仅当x>4,f(x)>x2-4x+5,求f(x)的解](/uploads/image/z/13866775-7-5.jpg?t=%E5%B7%B2%E7%9F%A5%E4%B8%89%E6%AC%A1%E5%87%BD%E6%95%B0f%28x%29%3Dx3%2Bax2%2Bbx%2Bc%E5%9C%A8%28-%E2%88%9E%2C-1%29.%282.%2B%E2%88%9E%29%E4%B8%8A%E5%8D%95%E8%B0%83%E9%80%92%E5%A2%9E%2C%E5%9C%A8%EF%BC%88-1%2C2%EF%BC%89%E4%B8%8A%E5%8D%95%E8%B0%83%E9%80%92%E5%B7%B2%E7%9F%A5%E4%B8%89%E6%AC%A1%E5%87%BD%E6%95%B0f%28x%29%3Dx3%2Bax2%2Bbx%2Bc%E5%9C%A8%28-%E2%88%9E%2C-1%29.%282.%2B%E2%88%9E%29%E4%B8%8A%E5%8D%95%E8%B0%83%E9%80%92%E5%A2%9E%2C%E5%9C%A8%28-1%2C2%29%E4%B8%8A%E5%8D%95%E8%B0%83%E9%80%92%E5%87%8F%2C%E8%8B%A5%E5%BD%93%E4%B8%94%E4%BB%85%E5%BD%93x%EF%BC%9E4%2Cf%EF%BC%88x%EF%BC%89%EF%BC%9Ex2-4x%2B5%2C%E6%B1%82f%EF%BC%88x%EF%BC%89%E7%9A%84%E8%A7%A3)
已知三次函数f(x)=x3+ax2+bx+c在(-∞,-1).(2.+∞)上单调递增,在(-1,2)上单调递已知三次函数f(x)=x3+ax2+bx+c在(-∞,-1).(2.+∞)上单调递增,在(-1,2)上单调递减,若当且仅当x>4,f(x)>x2-4x+5,求f(x)的解
已知三次函数f(x)=x3+ax2+bx+c在(-∞,-1).(2.+∞)上单调递增,在(-1,2)上单调递
已知三次函数f(x)=x3+ax2+bx+c在(-∞,-1).(2.+∞)上单调递增,在(-1,2)上单调递减,若当且仅当x>4,f(x)>x2-4x+5,求f(x)的解析式
已知三次函数f(x)=x3+ax2+bx+c在(-∞,-1).(2.+∞)上单调递增,在(-1,2)上单调递已知三次函数f(x)=x3+ax2+bx+c在(-∞,-1).(2.+∞)上单调递增,在(-1,2)上单调递减,若当且仅当x>4,f(x)>x2-4x+5,求f(x)的解
f'(x)=3x²+2ax+b
因为函数在(-∞,-1)∪(2,+∞)上单调递增,在(-1,2)上单调递减.
所以函数在x=-1和2时有极值,即导函数f'(x)在x=-1和2时为0.
则导函数为:f'(x)=3(x+1)(x-2)=3x²-3x-6,f''(x)=6x-3.
对比系数得:a=-3/2;b=-6.
则f(x)=x³-3/2·x²-6x+c.
设h(x)=x²-4x+5,即h'(x)=2x-4,h''(x)=2.
f(4)=16+c;h(4)=5.
有16+c=5,c=-11.
f(x)、h(x)在x>4时,都为单调递增.
f(x)和h(x)在x=4相交,而f'(4)=30>h'(4)=4,f''(4)=21>h''(4)=2.
所以f(x)增加幅度比h(x)大,即f(x)在h(x)上方.
所以有x>4时,f(x)>x²-4x+5.
总之:f(x)=x³-3/2·x²-6x-11.
求导。
g(x)=3x^2+2ax+b,g(-1)=g(2)=0
b=-12.a=4.5