求助,这道数学题怎么做?已知函数f(x)=a2^x+b3^x,其中常数a,b满足abf(x)时的X的取值范围.

Question

求助,这道数学题怎么做?
已知函数f(x)=a2^x+b3^x,其中常数a,b满足abf(x)时的X的取值范围.

Answer 1

f(x+1)-f(x)=a2^(x+1)+b3^(x+1)-a2^x-b3^x=2*a2^x+3*b3^x-a2^x-b3^x=a2^x+2b3^x
所以f(x+1)>f(x) f(x+1)-f(x)>0 a2^x+2b3^x>0
若a0 则有 2b3^x>-a2^x=|a|2^x (1.5)^x>|a|/2b (1.5)^x>|a|/2b x>log(1.5)(|a|/2b)
若a>0 ,b-2b3^x=2|b|3^x (2/3)^x>2|b|/a x>log(2/3)(2|b|/a)