作业帮已知二次函数f(x)满足f(x+1)-f(x)=2x且f(0)=1,设g(t)=f(2t+a),t∈[-1,1],求g(t)的求g(t)的最大值
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![已知二次函数f(x)满足f(x+1)-f(x)=2x且f(0)=1,设g(t)=f(2t+a),t∈[-1,1],求g(t)的求g(t)的最大值](/uploads/image/z/10947225-57-5.jpg?t=%E5%B7%B2%E7%9F%A5%E4%BA%8C%E6%AC%A1%E5%87%BD%E6%95%B0f%28x%29%E6%BB%A1%E8%B6%B3f%28x%2B1%29-f%28x%29%3D2x%E4%B8%94f%280%29%3D1%2C%E8%AE%BEg%28t%29%3Df%282t%2Ba%29%2Ct%E2%88%88%5B-1%2C1%5D%2C%E6%B1%82g%28t%29%E7%9A%84%E6%B1%82g%28t%29%E7%9A%84%E6%9C%80%E5%A4%A7%E5%80%BC)
已知二次函数f(x)满足f(x+1)-f(x)=2x且f(0)=1,设g(t)=f(2t+a),t∈[-1,1],求g(t)的求g(t)的最大值
已知二次函数f(x)满足f(x+1)-f(x)=2x且f(0)=1,设g(t)=f(2t+a),t∈[-1,1],求g(t)的
求g(t)的最大值
已知二次函数f(x)满足f(x+1)-f(x)=2x且f(0)=1,设g(t)=f(2t+a),t∈[-1,1],求g(t)的求g(t)的最大值
f(x)>2x+m
x^2-x+1>2x+m
x^2-3x+1-m>0
令g(x)=x^2-3x+1-m
可以得出当x0
m< - 3
(3)g(t)=f(2t+a)=4*t^2+(4a-2)*t+a^2-a+1
g(t)=(2t+a-1/2)^2+3/4
则g(t)的最大值只可能在g（-1）和g(1)中取
g(-1)=a^2-5a+7
g(1)=a^2+3a+3
当a>=1/2时
g(1)>=g(-1)
最大值为g(1)=a^2+3a+3
当a
二次函数f(x)满足 f(0)=1 ，
设 f(x) = ax^2 +bx +1
f(x+1)-f(x)= [a(x+1)^2 +b(x+1) +1]-[ax^2 +bx +1] = 2ax + a+b = 2x
a = 1 , b= -1
∴ f(x) = x^2 -x +1
g(t)=f(2t+1) = (2t+1)^2 -(2t+1) +1 = 4t^...
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二次函数f(x)满足 f(0)=1 ，
设 f(x) = ax^2 +bx +1
f(x+1)-f(x)= [a(x+1)^2 +b(x+1) +1]-[ax^2 +bx +1] = 2ax + a+b = 2x
a = 1 , b= -1
∴ f(x) = x^2 -x +1
g(t)=f(2t+1) = (2t+1)^2 -(2t+1) +1 = 4t^2 +2t +1 = (2t+1/2)^2 + 3/4
最小值：g(-1/4) = 3/4
g(-1) = 3 , g(1) = 7
∴ g(t)的最大值为：g（1）= 7
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