Question

函數(shù)f(x)的定義域是R，f(0)＝2，對任意x∈R，f(x)＋f′(x)>1，則不等式e^x·f(x)>e^x＋1的解集為______．

Answer 1

(0，＋∞)

構(gòu)造函數(shù)g(x)＝e^x·f(x)－e^x，因為g′(x)＝e^x·f(x)＋e^x·f′(x)－e^x＝e^x[f(x)＋f′(x)]－e^x>e^x－e^x＝0，所以g(x)＝e^x·f(x)－e^x為R上的增函數(shù)．又因為g(0)＝e⁰·f(0)－e⁰＝1，所以原不等式轉(zhuǎn)化為g(x)>g(0)，解得x>0.