已知函数f(x)=ax+b/x+c (a,b,c∈R)满足f(-1)=0,并且对x>0,0≤f(x)-1≤(x-1)^2/2x已知函数f(x)=ax+(b/x)+c (a,b,c∈R)满足f(-1)=0,并且对x>0,0≤f(x)-1≤((x-1)^2)/2x恒成立,求a,b,c
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![已知函数f(x)=ax+b/x+c (a,b,c∈R)满足f(-1)=0,并且对x>0,0≤f(x)-1≤(x-1)^2/2x已知函数f(x)=ax+(b/x)+c (a,b,c∈R)满足f(-1)=0,并且对x>0,0≤f(x)-1≤((x-1)^2)/2x恒成立,求a,b,c](/uploads/image/z/12556603-19-3.jpg?t=%E5%B7%B2%E7%9F%A5%E5%87%BD%E6%95%B0f%28x%29%3Dax%2Bb%2Fx%2Bc+%28a%2Cb%2Cc%E2%88%88R%29%E6%BB%A1%E8%B6%B3f%28-1%29%3D0%2C%E5%B9%B6%E4%B8%94%E5%AF%B9x%3E0%2C0%E2%89%A4f%28x%29-1%E2%89%A4%28x-1%29%5E2%2F2x%E5%B7%B2%E7%9F%A5%E5%87%BD%E6%95%B0f%28x%29%3Dax%2B%EF%BC%88b%2Fx%EF%BC%89%2Bc+%28a%2Cb%2Cc%E2%88%88R%29%E6%BB%A1%E8%B6%B3f%28-1%29%3D0%2C%E5%B9%B6%E4%B8%94%E5%AF%B9x%3E0%2C0%E2%89%A4f%28x%29-1%E2%89%A4%EF%BC%88%28x-1%29%5E2%EF%BC%89%2F2x%E6%81%92%E6%88%90%E7%AB%8B%2C%E6%B1%82a%2Cb%2Cc)
已知函数f(x)=ax+b/x+c (a,b,c∈R)满足f(-1)=0,并且对x>0,0≤f(x)-1≤(x-1)^2/2x已知函数f(x)=ax+(b/x)+c (a,b,c∈R)满足f(-1)=0,并且对x>0,0≤f(x)-1≤((x-1)^2)/2x恒成立,求a,b,c
已知函数f(x)=ax+b/x+c (a,b,c∈R)满足f(-1)=0,并且对x>0,0≤f(x)-1≤(x-1)^2/2x
已知函数f(x)=ax+(b/x)+c (a,b,c∈R)满足f(-1)=0,并且对x>0,0≤f(x)-1≤((x-1)^2)/2x恒成立,求a,b,c
已知函数f(x)=ax+b/x+c (a,b,c∈R)满足f(-1)=0,并且对x>0,0≤f(x)-1≤(x-1)^2/2x已知函数f(x)=ax+(b/x)+c (a,b,c∈R)满足f(-1)=0,并且对x>0,0≤f(x)-1≤((x-1)^2)/2x恒成立,求a,b,c
f(x) = ax + (b/x) + c
已知函数满足f(-1) = 0,则有
f(-1) = -a - b + c = 0 .(1)
考察右边的不等式,令x = 1,有
0 ≤ f(1) - 1 ≤ ((1-1)^2)/2
即 1 ≤ f(1) ≤ 1,因此
f(1) = a + b + c = 1 .(2)
联立(1),(2)
-a - b + c = 0
a + b + c = 1
首先得到
c = 1/2 .(3)
a + b = 1/2 .(4)
由于不等式中约定x > 0,因此将不等式两边乘以x,得到:
0 ≤ (ax^2 + cx + b) - x ≤ ((x-1)^2)/2
先看第一个不等式:
ax^2 + (c-1)x + b >= 0 (x > 0)
因此需满足(二次函数):
a > 0.(5)
(c-1)^2 - 4ab ≤ 0 .(6)
或(一次函数):
a = 0.(5')
b >= 0.(6')
c-1 >= 0.(7')
因为 c = 1/2,所以(5')~(7')不成立.
将(1')(2')代入(6),有
ab >= 1/16 .(7)
a + b = 1/2 .(4)
a > 0 .(5)
显然满足此条件的只能是
a = 1/4
b = 1/4
(如果要证明的话,将(4)代入(7)得到不等式a^2 - (1/2)a + 1/16