已知二次函数f(x)=ax^2+bx,f(x-1)为偶函数,集合A={X|f(x)=x}为单元素集合(1)求f(x)解析式(2)设函数g(x)=[f(x)-m]*e^x,若函数g(x)在x∈[-3,2]上单调,求实数m取值范围.
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![已知二次函数f(x)=ax^2+bx,f(x-1)为偶函数,集合A={X|f(x)=x}为单元素集合(1)求f(x)解析式(2)设函数g(x)=[f(x)-m]*e^x,若函数g(x)在x∈[-3,2]上单调,求实数m取值范围.](/uploads/image/z/3154623-15-3.jpg?t=%E5%B7%B2%E7%9F%A5%E4%BA%8C%E6%AC%A1%E5%87%BD%E6%95%B0f%28x%29%3Dax%5E2%2Bbx%2Cf%28x-1%29%E4%B8%BA%E5%81%B6%E5%87%BD%E6%95%B0%2C%E9%9B%86%E5%90%88A%3D%7BX%7Cf%28x%29%3Dx%7D%E4%B8%BA%E5%8D%95%E5%85%83%E7%B4%A0%E9%9B%86%E5%90%88%EF%BC%881%EF%BC%89%E6%B1%82f%28x%29%E8%A7%A3%E6%9E%90%E5%BC%8F%EF%BC%882%EF%BC%89%E8%AE%BE%E5%87%BD%E6%95%B0g%28x%29%3D%5Bf%28x%29-m%5D%2Ae%5Ex%2C%E8%8B%A5%E5%87%BD%E6%95%B0g%28x%29%E5%9C%A8x%E2%88%88%5B-3%2C2%5D%E4%B8%8A%E5%8D%95%E8%B0%83%2C%E6%B1%82%E5%AE%9E%E6%95%B0m%E5%8F%96%E5%80%BC%E8%8C%83%E5%9B%B4.)
已知二次函数f(x)=ax^2+bx,f(x-1)为偶函数,集合A={X|f(x)=x}为单元素集合(1)求f(x)解析式(2)设函数g(x)=[f(x)-m]*e^x,若函数g(x)在x∈[-3,2]上单调,求实数m取值范围.
已知二次函数f(x)=ax^2+bx,f(x-1)为偶函数,集合A={X|f(x)=x}为单元素集合
(1)求f(x)解析式
(2)设函数g(x)=[f(x)-m]*e^x,若函数g(x)在x∈[-3,2]上单调,求实数m取值范围.
已知二次函数f(x)=ax^2+bx,f(x-1)为偶函数,集合A={X|f(x)=x}为单元素集合(1)求f(x)解析式(2)设函数g(x)=[f(x)-m]*e^x,若函数g(x)在x∈[-3,2]上单调,求实数m取值范围.
平方之类的我不会用键盘打,我告诉你方法.
因为f(x-1)为偶函数,所以由偶函数f(-x)=-f(x)也就是
f[-(x-1)]=-f(x-1)
a(1-x)^2+b(1-x)=-[a(x-1)^2+b(x-1)]
又因为A为单元素,单元素的意思是f(x)=x 即ax^2+bx=x的方程有一个根.
(1) f(x-1) = a(x - 1)² + b(x - 1) = ax² - 2ax + a + bx - b = ax² + (b - 2a)x + a - b
f(x1)为偶函数, b - 2a = 0, b = 2a
f(x) = ax² + 2ax
f(x) = x, ax² + 2ax = x, ax...
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(1) f(x-1) = a(x - 1)² + b(x - 1) = ax² - 2ax + a + bx - b = ax² + (b - 2a)x + a - b
f(x1)为偶函数, b - 2a = 0, b = 2a
f(x) = ax² + 2ax
f(x) = x, ax² + 2ax = x, ax² + (2a-1)x = 0
x[ax + (2a-1)] = 0
x₁ = 0, x₂ = 1-2a
集合A为单元素集合, 1-2a = 0, a = 1/2
f(x) = x²/2 + x
(2) g(x) = (x²/2 + x -m)e^x
g'(x) = (x + 1)e^x + (x²/2 + x -m)e^x = (x²/2 + 2x + 1 -m)e^x = 0
x² + 4x + 2(1 -m) = 0
∆ = 16-4*2(1-m) = 8+8m
x₁₂ = [-4 ± √(8+8m)]/2 = -2 ± √(2+2m)
x₁ = -2 - √(2+2m), x₂ = -2 + √(2+2m)
要使g(x)在x∈[-3,2]上单调, 有三种可能:
(i) x₁ ≥ 2
-2 - √(2+2m) ≥ 2
√(2+2m) ≤ -4, 不可能
(ii) x₂ ≤ -3
-2 + √(2+2m) ≤ -3
√(2+2m) ≤ -1, 不可能
(iii) x₁ ≤ -3且x₂ ≥ 2
-2 - √(2+2m) ≤ -3
√(2+2m) ≥ 1
m ≥ -1/2 (a)
-2 + √(2+2m) ≥ 2
√(2+2m) ≥ 4
m ≥ 7 (b)
结合(a)(b): m ≥ 7
收起