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設y=f(x)=x+lnx 定義域為:x>0 ∵y‘=1+1/x >0 ∴ f(x) 單調增 ∵ f(e^(-3))=e^(-3)-30 所以 在 (e^(-3),e) 區間 函數f(x)有一個零點 即 方程x+lnx=0實根的個數為1
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