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导数已是新课标教学的重要内容之一,每年高考卷几乎都有一道与导数有关的压轴题,反映了导数的重要性,也使数学方法更加丰富,新课标试卷将导数与传统的不等式结合,是一种新颖的命题模式,体现了导数作为工具,用于分析和解决一些函数问题很方便,以下介绍几道运用导数方法证明不等式的例题。例1.(2013年全国新课标Ⅱ卷,理科21题第2问)已知函数f(x)=e~x-ln(x+m)当m≤2时,证明f(x)>0解:∵m≤2,∴ln(x+m)≤ln(x+2)(?)e~x-ln(x+m)≥e~x-ln(x+2)(1)又∵e~x>x+1,∴e~x-ln(x+2)≥x+1-ln(x+2)(2)由(1)(2)知e~x-ln(x+m)≥x+1-ln(x+2)即:f(x)≥x+1-ln(x+2)设h(x)=x+1-ln(x+2)∴h′(x)=1-1/x+2=x+1/x+2当x∈(-2,-1)时,h′(x)<0,当x∈(-1+∞)时,
Derivatives is one of the important content of the new curriculum teaching, the annual college entrance examination almost every year with a derivative of the finale question, reflecting the importance of derivatives, but also to enrich the mathematical method, the new curriculum standard test paper derivatives and traditional Inequality combination is a novel propositional model that embodies the derivative as a tool for analyzing and solving some of the function problems is very convenient, the following describes a few examples using the derivative method to prove inequalities. Example 1. The function f (x) = e ~ x-ln (x + m) is known to be f (x)> m 0 Solution: ∵m≤2, ∴ln (x + m) ≤ln (x + 2) (?) E ~ x-ln (x + m) ≥e ~ x-ln (x + 2) (2) From (1) (2) we know e ~ x - ln (x + 2) Let x (x) = x + 1-ln (x + 2) ∴h ’(x + 2) when x∈ (-1, -1), h ’(x) <0, when x∈ (-1 + ∞)