证明不等式《高等数学》
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证明不等式《高等数学》
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![证明不等式《高等数学》](/uploads/image/z/16044892-52-2.jpg?t=%E8%AF%81%E6%98%8E%E4%B8%8D%E7%AD%89%E5%BC%8F%E3%80%8A%E9%AB%98%E7%AD%89%E6%95%B0%E5%AD%A6%E3%80%8B)
ln(1+0)=0,
假设x>0,prove x-ln(1+x)>0,let f(x)=x-ln(1+x),f(0) = 0,f'(x) = 1-[1/(1+x)] = x/(1+x)>0,so f(x) is increasing in [0,+正无穷),so f(x)>0,so x>ln(1+x).
prove ln(1+x) - [x/(1+x)] >0,let g(x) = ln(1+x) - [x/(1+x)] = ln(1+x) - [1-1/(1+x))],
so g(0) = 0,g'(x) = 1/(1+x) -1/[(1+x)^2] = x/[(1+x)^2] >0,where x>0,
so g(x) is increasing in [0,+正无穷),so g(x)>0 for x>0,so ln(1+x) > x/(1+x)
假设x>0,prove x-ln(1+x)>0,let f(x)=x-ln(1+x),f(0) = 0,f'(x) = 1-[1/(1+x)] = x/(1+x)>0,so f(x) is increasing in [0,+正无穷),so f(x)>0,so x>ln(1+x).
prove ln(1+x) - [x/(1+x)] >0,let g(x) = ln(1+x) - [x/(1+x)] = ln(1+x) - [1-1/(1+x))],
so g(0) = 0,g'(x) = 1/(1+x) -1/[(1+x)^2] = x/[(1+x)^2] >0,where x>0,
so g(x) is increasing in [0,+正无穷),so g(x)>0 for x>0,so ln(1+x) > x/(1+x)