关于介值定理的专题讨论

$f命题:$设$fleft( x ight) in Cleft[ {a,b} ight]$,则存在$xi  in left[ {a,b} ight]$,使得$fleft( xi   ight) = frac{{{c_1}fleft( {{x_1}} ight) +  cdots  + {c_n}fleft( {{x_n}} ight)}}{{{c_1} +  cdots  + {c_n}}}$,其中${c_i} > 0,a < {x_1} < ... < {x_n} < b$

方法一

$f命题:$设$fleft( x ight) in Cleft( {a,b} ight)$,且$lim limits_{x o egin{array}{*{20}{c}}{{a^ + }}end{array}} fleft( x ight) = A,lim limits_{x o egin{array}{*{20}{c}}{{b^ - }}end{array}} fleft( x ight) = B$,则对任意$c in left( {A,B} ight)$,存在$xi  in left( {a,b} ight)$,使得$fleft( xi   ight) = c $

方法一

$f命题:$设$fleft( x ight)$在$left[ {0, + infty } ight)$上有连续导数,且$f'left( x ight) ge k > 0,fleft( 0 ight) < 0$,则$fleft( x ight)$在$left( {0, + infty } ight)$上有且仅有一个零点

方法一

$f命题:$开普勒$fleft( {Kepler} ight)$方程$x = varepsilon sin x + aleft( {0 < varepsilon  < 1} ight)$只有唯一实根

方法一

$f命题:$设$f(x)$在$(a,b)$内二阶可导,$c$为$(a,b)$内一点,满足$f''left( c ight) e 0$,则存在不同的点${x_1},{x_2} in left( {a,b} ight)$,使得[frac{{fleft( {{x_2}} ight) - fleft( {{x_1}} ight)}}{{{x_2} - {x_1}}} = f'left( c ight)]

方法一

$f命题:$

 






原文地址:https://www.cnblogs.com/ly142857/p/3676724.html