作业帮设函数f(x)=a^x+3a(a>0,a不等于1),g(x)=loga (x-a),若对任意x属于[a+2,a+3]时,恒有|f-1(x)+g(x)|
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![设函数f(x)=a^x+3a(a>0,a不等于1),g(x)=loga (x-a),若对任意x属于[a+2,a+3]时,恒有|f-1(x)+g(x)|](/uploads/image/z/3724226-26-6.jpg?t=%E8%AE%BE%E5%87%BD%E6%95%B0f%28x%29%3Da%5Ex%2B3a%28a%3E0%2Ca%E4%B8%8D%E7%AD%89%E4%BA%8E1%29%2Cg%28x%29%3Dloga+%28x-a%29%2C%E8%8B%A5%E5%AF%B9%E4%BB%BB%E6%84%8Fx%E5%B1%9E%E4%BA%8E%5Ba%2B2%2Ca%2B3%5D%E6%97%B6%2C%E6%81%92%E6%9C%89%7Cf-1%28x%29%2Bg%28x%29%7C)
设函数f(x)=a^x+3a(a>0,a不等于1),g(x)=loga (x-a),若对任意x属于[a+2,a+3]时,恒有|f-1(x)+g(x)|
设函数f(x)=a^x+3a(a>0,a不等于1),g(x)=loga (x-a),若对任意x属于[a+2,a+3]时,恒有|f-1(x)+g(x)|
设函数f(x)=a^x+3a(a>0,a不等于1),g(x)=loga (x-a),若对任意x属于[a+2,a+3]时,恒有|f-1(x)+g(x)|
设函数f(x)=a^x+3a(a>0,a不等于1),g(x)=loga (x-a),若对任意x属于[a+2,a+3]时,恒有|f-1(x)+g(x)|0,a不等于1)
∴f^(-1)(x)=log(a,x-3a) (x>3a)
∵g(x)=log(a,x-a) (x>a)
f^(-1)(x)+g(x)=log(a,x-3a)+log(a,x-a)=log(a,(x-3a)(x-a)) (x>3a)
∵对任意x属于[a+2,a+3]时,恒有|f-1(x)+g(x)|a+√(a^2+1/a)≤3≤a+√(a^2+a)显然也无解
当0a-√(a^2+a)≤3≤a-√(a^2+1/a)
显然也无解.
或者,2a+√(a^2+a)≤a+2≤2a+√(a^2+1/a)
2a+√(a^2+a)≤a+2==>√(a^2+a)≤2-a==>则0＜a≤4/5
a+2≤2a+√(a^2+1/a)==>2-a≤√(a^2+1/a)==>4a^2-4a+1≥0==>(2a-1)^2≥0∴a+2≤2a+√(a^2+1/a)恒成立
∴0＜a≤4/5
2a+√(a^2+a)≤a+3≤2a+√(a^2+1/a)
2a+√(a^2+a)≤a+3==>0＜a≤9/7
a+3≤2a+√(a^2+1/a)==>√(a^2+1/a)≥3-a==>1/a+6a-9≥0
得,(9-√57)/12＜a≤(9+√57)/12
∵0
设函数f(x)=a^x+3a(a>0,a不等于1),g(x)=loga (x-a),若对任意x属于[a+2,a+3]时，恒有|f-1(x)+g(x)|<=1成立，求实数a的取值范围。 f-1（x）为反函数