设函数f(x)=(3x+4)/(x^2+1,g(x)=6a^2/x+a a>1/3 若对任意x0∈[0,a]总存在相应的x1,x2∈[0,a],使得g(x设函数f(x)=(3x+4)/(x^2+1),g(x)=(6a^2)/(x+a) a>1/3 若对任意x0∈[0,a]总存在相应的x1,x2∈[0,a],使得g(x1)《f
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![设函数f(x)=(3x+4)/(x^2+1,g(x)=6a^2/x+a a>1/3 若对任意x0∈[0,a]总存在相应的x1,x2∈[0,a],使得g(x设函数f(x)=(3x+4)/(x^2+1),g(x)=(6a^2)/(x+a) a>1/3 若对任意x0∈[0,a]总存在相应的x1,x2∈[0,a],使得g(x1)《f](/uploads/image/z/5411625-33-5.jpg?t=%E8%AE%BE%E5%87%BD%E6%95%B0f%28x%29%3D%283x%2B4%29%2F%28x%5E2%2B1%2Cg%28x%29%3D6a%5E2%2Fx%2Ba+a%3E1%2F3+%E8%8B%A5%E5%AF%B9%E4%BB%BB%E6%84%8Fx0%E2%88%88%5B0%2Ca%5D%E6%80%BB%E5%AD%98%E5%9C%A8%E7%9B%B8%E5%BA%94%E7%9A%84x1%2Cx2%E2%88%88%5B0%2Ca%5D%2C%E4%BD%BF%E5%BE%97g%EF%BC%88x%E8%AE%BE%E5%87%BD%E6%95%B0f%28x%29%3D%283x%2B4%29%2F%28x%5E2%2B1%EF%BC%89%2Cg%28x%29%3D%EF%BC%886a%5E2%EF%BC%89%2F%EF%BC%88x%2Ba%EF%BC%89+a%3E1%2F3+%E8%8B%A5%E5%AF%B9%E4%BB%BB%E6%84%8Fx0%E2%88%88%5B0%2Ca%5D%E6%80%BB%E5%AD%98%E5%9C%A8%E7%9B%B8%E5%BA%94%E7%9A%84x1%2Cx2%E2%88%88%5B0%2Ca%5D%2C%E4%BD%BF%E5%BE%97g%EF%BC%88x1%EF%BC%89%E3%80%8Af)
设函数f(x)=(3x+4)/(x^2+1,g(x)=6a^2/x+a a>1/3 若对任意x0∈[0,a]总存在相应的x1,x2∈[0,a],使得g(x设函数f(x)=(3x+4)/(x^2+1),g(x)=(6a^2)/(x+a) a>1/3 若对任意x0∈[0,a]总存在相应的x1,x2∈[0,a],使得g(x1)《f
设函数f(x)=(3x+4)/(x^2+1,g(x)=6a^2/x+a a>1/3 若对任意x0∈[0,a]总存在相应的x1,x2∈[0,a],使得g(x
设函数f(x)=(3x+4)/(x^2+1),g(x)=(6a^2)/(x+a) a>1/3 若对任意x0∈[0,a]总存在相应的x1,x2∈[0,a],使得g(x1)《f(x0)《g(x2)成立 求实数a取值
设函数f(x)=(3x+4)/(x^2+1,g(x)=6a^2/x+a a>1/3 若对任意x0∈[0,a]总存在相应的x1,x2∈[0,a],使得g(x设函数f(x)=(3x+4)/(x^2+1),g(x)=(6a^2)/(x+a) a>1/3 若对任意x0∈[0,a]总存在相应的x1,x2∈[0,a],使得g(x1)《f
设函数f(x)=(3x+4)/(x^2+1),g(x)=(6a^2)/(x+a) a>1/3 若对任意x0∈[0,a]总存在相应的x1,x2∈[0,a],使得g(x1)《f(x0)《g(x2)成立 求实数a取值
解析:∵函数f(x)=(3x+4)/(x^2+1),g(x)=(6a^2)/(x+a) a>1/3
令f’(x)=(3-8x-6x^2)/(x^2+1)^2=0==>x1=1/3,x2=-3(舍)
∴函数f(x)在x=1/3处取极大值f(1/3)=9/2
G’(x)=-(6a^2)/(x+a)^20时,单调减
∵对任意x0∈[0,a]总存在相应的x1,x2∈[0,a],使得g(x1)