已知二次函数f(x)=-x²+(2t-1)x+(t²+1) ①若f(x)在(-无穷大,-1]上为增已知二次函数f(x)=-x²+(2t-1)x+(t²+1)①若f(x)在(-无穷大,-1]上为增函数,求t的取
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![已知二次函数f(x)=-x²+(2t-1)x+(t²+1) ①若f(x)在(-无穷大,-1]上为增已知二次函数f(x)=-x²+(2t-1)x+(t²+1)①若f(x)在(-无穷大,-1]上为增函数,求t的取](/uploads/image/z/6160951-55-1.jpg?t=%E5%B7%B2%E7%9F%A5%E4%BA%8C%E6%AC%A1%E5%87%BD%E6%95%B0f%EF%BC%88x%EF%BC%89%EF%BC%9D-x%26%23178%3B%EF%BC%8B%EF%BC%882t-1%EF%BC%89x%EF%BC%8B%EF%BC%88t%26%23178%3B%EF%BC%8B1%EF%BC%89+%E2%91%A0%E8%8B%A5f%EF%BC%88x%EF%BC%89%E5%9C%A8%EF%BC%88-%E6%97%A0%E7%A9%B7%E5%A4%A7%2C-1%EF%BC%BD%E4%B8%8A%E4%B8%BA%E5%A2%9E%E5%B7%B2%E7%9F%A5%E4%BA%8C%E6%AC%A1%E5%87%BD%E6%95%B0f%EF%BC%88x%EF%BC%89%EF%BC%9D-x%26%23178%3B%EF%BC%8B%EF%BC%882t-1%EF%BC%89x%EF%BC%8B%EF%BC%88t%26%23178%3B%EF%BC%8B1%EF%BC%89%E2%91%A0%E8%8B%A5f%EF%BC%88x%EF%BC%89%E5%9C%A8%EF%BC%88-%E6%97%A0%E7%A9%B7%E5%A4%A7%2C-1%EF%BC%BD%E4%B8%8A%E4%B8%BA%E5%A2%9E%E5%87%BD%E6%95%B0%2C%E6%B1%82t%E7%9A%84%E5%8F%96)
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已知二次函数f(x)=-x²+(2t-1)x+(t²+1) ①若f(x)在(-无穷大,-1]上为增已知二次函数f(x)=-x²+(2t-1)x+(t²+1)①若f(x)在(-无穷大,-1]上为增函数,求t的取
已知二次函数f(x)=-x²+(2t-1)x+(t²+1) ①若f(x)在(-无穷大,-1]上为增
已知二次函数f(x)=-x²+(2t-1)x+(t²+1)
①若f(x)在(-无穷大,-1]上为增函数,求t的取值范围
②设f(x)的最大值为g(t),求g(t)的表达式,并求g(t)的最小值
已知二次函数f(x)=-x²+(2t-1)x+(t²+1) ①若f(x)在(-无穷大,-1]上为增已知二次函数f(x)=-x²+(2t-1)x+(t²+1)①若f(x)在(-无穷大,-1]上为增函数,求t的取
1)当x≤-1时为增函数,则对称轴需在区间右边
即(2t-1)/2≥-1
得:t≥-1/2
2)配方:f(x)=-[x-(t-1/2)]^2+t^2+1+(t-1/2)^2=-[x-(t-1/2)]^2+2t^2-t+5/4
最大值g(t)=2t^2-t+5/4
对g(t)与方:g(t)=2(t-1/4)^2+5/4-1/8=2(t-1/4)^2+9/8
即g(t)的最小值为9/8