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Last edited by hbghlyj 2020-2-9 20:42等价于$f(x)=\frac1{1+x^\alpha},x\in\mathbf R^+$的割线斜率$\in$[-1,1],
等价于$f'(x)=-\frac{\alpha x^{\alpha -1}}{\left(x^{\alpha }+1\right)^2}\in[-1,1],\forall x\in\mathbf R^+$.
$f''(x)=\frac{\alpha x^{\alpha -2} \left(-\alpha +\alpha x^{\alpha }+x^{\alpha }+1\right)}{\left(x^{\alpha }+1\right)^3}$,
所以$\alpha\in[-a,-1]\cup[1,a]$
其中a=3.717430610620688...为$-\frac{\alpha ^2-1}{4 \alpha }\left(\frac{\alpha +1}{\alpha -1}\right)^{\frac1\alpha } $的零点 |
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hbghlyj
Posted 2020-2-9 20:26
