已知函数y=f(x),编程实现输入一个x值,输出y值

（1）取y=0,于是f(x)=f(x)*f(0),对任意的x属于R,我们知道f(0)=1可以取这样的f(x)=e^x,顺便可以验证一下正确性,f(0)=1(2)①当x0,取y=-x,于是f(x-x)=

intsgn(intx){if(x

令x＝y＝0得2f(0)=2f^2(0),于是f(0)=0.（因为f(0)不为1）.再令x＝0得f(y)+f(－y)=2f(0)f(y)=0,因此f(－y)＝－f(y),f是奇函数.显然有F(－x)=

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(1)令x=y=0f(x+y)=f(x)f(y)-f(x)-f(y)+2变为f(0)=f(0)^2-2f(0)+2f(0)^2-3f(0)+2=0(f(0)-1)(f(0)-2)=0f(0)=1或f(

f(-3)=f(-3/2-3/2)=f(-3/2)+f(-3/2)=af(-3/2)=a/2

y'=2f'（2x),y''=2x2f''（2x).这是复合函数求导原则,举例f(a(X））的导数为f'(a(X))乘以a'(X)

令y=-xf(x-x)=f(0)=f(x)+f(-x)f(0+0)=f(0)+f(0)=0故f(x)+f(-x)=0从而f(x)=-f(-x)奇函数得证f(3)=-f(-3)=-af(6)=f(3)+

1.令x=y=0f(0)=f(0)+f(0)=2f(0),f(0)=0令y=-xf(0)=f(x)+f(-x)=0f(-x)=-f(x)得证2.令x>yf(x-y)

1,令y=-xf(0)=f(x)+f(-x)2,f(x)+f(-x)=0f(x)=-f(-x)f(3)=-af(6)=f(3)+f(3)=-2af(12)=2f(6)=-4a3,f(x)=-f(-x)

令x=y=0,则f(0)=0.令y=-x,则f(0)=f(x)+f(-x),则f(x)在R上为奇函数.f(x+y)=f(x)+f(y),有f(x-y)=f(x)+f(-y)=f(x)-f(y).

设a=xy,b=x+y.f(xy,x+y)=x^2+y^2+2xy-2xy=(x+y)^2-2xy把a,b带f(a,b)=b^2-2a所以f(x,y)=y^2-2x同理f(x+y,xy)=x^2+y^

（1）令XY为0,则f(x+y)=f(x)+f(y)f(0)=f(0)+f(0)所以f(0)=0再令Y=-X所以f(x-x)=f(x)+f(-x)所以f(x)=-f(-x)即f(x)是奇函数（2）因f

publicclassDemo{publicstaticvoidcalc(doublex){doubley=0;if(x0&&x

//if()实现#include<iostream.h>#include<math.h>voidmain(){    doublex,y

C

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编程喜欢不好.不要省略{}有错误提示吗?再问：上面说intx;floaty;这里错了再答：总有错误代码的吧

x=linspace(1,5,1000);y=linspace(2,15,1000);f=x.^2+x.*y;fmin=min(f)fmax=max(f)fmin=3fmax=100再问：我对matl

x3+x=0则x（x2+1）=0在实数范围内只有x=0才是零点.