二西安毛子律技大学2xen(x-1) + ax + b例3. 设f(x)= limen(x-1) +1n-0试确定常数a,b使,f(x)处处可导,并求f'(x)ax+b,x<1解: f(x)=^ (a+b+1),x=1x2.x>1x<1时,f(x)=a; x>1时,f'(x)=2x利用f(x)在x=1处可导，得α+b =1=(α+b+1)f(1t)= f(1)= f(1)即a= 2f'(1) = f*(1)6
6 例3. 设 试确定常数 a , b 使 f (x) 处处可导,并求 解: f (x) = ax + b , x 1 ( 1), x = 1 2 1 a+ b + , x 1 2 x x 1时, f (x) = a ; x 1时，f (x) = 2x. (1) (1) − + f = f 利 用 f (x)在 x = 1处可导， 得 即 a + b = 1 ( 1) 2 1 = a + b + a = 2 f f f (1 ) (1 ) (1) + − = =
西安毛子律技枝大学=ax+b,x<1f(x)=3(a+ b + 1),x=1x2,x>1x<1时, f'(x)=a, x>1时,f'(x)=2xf'(1) = 2 a=2, b=-1,x≤1[2,f(x)=2x,x>17
7 a = 2, b = −1, f (1) = 2 = 2 , 1 2 , 1 ( ) x x x f x f (x) = ax + b , x 1 ( 1), x = 1 2 1 a+ b + , x 1 2 x x 1时, f (x) = a , x 1时，f (x) = 2x
三西安毛子律技大学2例4 设 f(x) 可微且 f(x)>0,(0)=f(0)=1, 求lim f(x)解 f(x)可微必连续，从而limf(x)=f(O)=1,1Lin f(x)lim=ln f(x)=el=elim f(x)* = limex=er>0xx→0x->0In f(x)In[1+ f(x) -1]f(x)-1limlimlimx->0x-0x->0xxxf(x)-f(0)= f'(0)= 1,- limx-0x->0
解 例4 设 f x( ) 可微且 f x f f ( ) 0, (0) (0) 1, = = 求 1 0 lim ( ) . x x f x → f x( ) 可微必连续, 0 lim ( ) (0) 1, x f x f → 从而 = = 1 1 ln ( ) 0 0 lim ( ) lime f x x x x x f x → → = 0 1 lim ln ( ) e x f x → x = 0 ln ( ) lim x f x → x = 0 ( ) 1 lim x f x → x − = 0 ( ) (0) lim x 0 f x f → x − = − = = f (0) 1,1 = = e e. 0 ln[1 ( ) 1] lim x f x → x + −
西安毛子律枝大学-例5 对任意正数x,y有 f(xy)= f(x)+ f(y)且 f'(l)=1.证明f(x)在(O,+o)内可导.证 f(11)= f(1)+ f(1), :: f(1)=0, Vxe(0,+),有f[x·(1+^)]- f(x)f(x+h)-f(x)f'(x)= lim ==lim4hhh->0h-→0hf(1+f(x)+ f(1+f(x)f(1xxx= limlimlim-hhhh-0h->0h->0xx1'(1)存在xx
例5 对任意正数 x y, 有 f x y f x f y ( ) ( ) ( ) = + 且 f (1) 1, = 证明 f x( ) 在 (0, ) + 内可导. 证 f f f (1 1) (1) (1), = + = f (1) 0, + x (0, ),有 0 ( ) ( ) ( ) limh f x h f x f x → h + − = 0 [ (1 )] ( ) lim h h f x f x x → h + − = 0 ( ) (1 ) ( ) lim h h f x f f x x → h + + − = 0 (1 ) lim h h f x → h + = 0 (1 ) (1) 1 lim h h f f x h x x → + − = 1 f (1) x = 1 . x = 存在
西安毛子律枝大学2x3,x≥1,例6 设函数 f(x)=3则f(x)在x=1处（x2, x<1.（A）可导且f(I)=2（B）左导数存在但右导数不存在（C）右导数存在但左导数不存在（D）左、右导数都不存在222x2解 x>1时，f(x)=f'(x)X33x<1 时， f(x)=x2, f(x)=(x22xf(2)=2x2, =2, f(2) X: f'(2) = 2 = 2
例6 设函数 3 2 2 , 1, ( ) 3 , 1. x x f x x x = 则 f x( ) 在 x =1 处（ ）. （A）可导且 f (1) 2 = （B）左导数存在但右导数不存在 （C）右导数存在但左导数不存在（D）左、右导数都不存在 解 x 1 2 3 ( ) , 3 时， f x x = 2 3 2 ( ) 2 , 3 f x x x = = x 1 2 时， f x x ( ) , = ( ) 2 f x x x ( ) 2 , = = 2 1 (2) 2 2, x f x + = = = 1 (2) 2 2, x f x − = = = = f (2) 2, A