设F(x)=f(x)+mg(x)+1-m-m^2,且|F(x)|在[0,1]上单调递增求m取值范围~f(x)=x^2,g(x)=x-1
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![设F(x)=f(x)+mg(x)+1-m-m^2,且|F(x)|在[0,1]上单调递增求m取值范围~f(x)=x^2,g(x)=x-1](/uploads/image/z/12350402-26-2.jpg?t=%E8%AE%BEF%28x%29%3Df%28x%29%2Bmg%28x%29%2B1-m-m%5E2%2C%E4%B8%94%7CF%28x%29%7C%E5%9C%A8%5B0%2C1%5D%E4%B8%8A%E5%8D%95%E8%B0%83%E9%80%92%E5%A2%9E%E6%B1%82m%E5%8F%96%E5%80%BC%E8%8C%83%E5%9B%B4%7Ef%28x%29%3Dx%5E2%2Cg%28x%29%3Dx-1)
设F(x)=f(x)+mg(x)+1-m-m^2,且|F(x)|在[0,1]上单调递增求m取值范围~f(x)=x^2,g(x)=x-1
设F(x)=f(x)+mg(x)+1-m-m^2,且|F(x)|在[0,1]上单调递增
求m取值范围~
f(x)=x^2,g(x)=x-1
设F(x)=f(x)+mg(x)+1-m-m^2,且|F(x)|在[0,1]上单调递增求m取值范围~f(x)=x^2,g(x)=x-1
函数f(x)和g(x)都没有,让哪路神仙帮你去啊?
F(x)=x²-m(x-1)+1-m-m²=x²-mx+1-m²
△=m²-4(1-m²)=5m²-4
△<0时-2√5/5
|F(x)|=x²-mx+1-m²单调递增区间为[m/2,+∞)
...
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F(x)=x²-m(x-1)+1-m-m²=x²-mx+1-m²
△=m²-4(1-m²)=5m²-4
△<0时-2√5/5
|F(x)|=x²-mx+1-m²单调递增区间为[m/2,+∞)
m/2≤0
m≤0
故-2√5/5
|F(x)| =0两根x1=[m-√(5m²-4)]/2,x2=[m+√(5m²-4)]/2,中间顶点横坐标m/2
|F(x)| 单调递增区间为[x1,m/2]和[x2,+∞)
x1≤0且m/2≥1或x2≤0
求解得m≥2或-1≤m≤1
故m≥2或2√5/5≤m≤1或-1≤m≤-2√5/5
因此m≥2或2√5/5≤m≤1或-1≤m≤0
知道答案,但是不明白为何要考虑
△<0时-2√5/5
|F(x)|=x²-mx+1-m²单调递增区间为[m/2,+∞)
m/2≤0
m≤0
故-2√5/5
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