f(x)在x=0处可导且f'(0)=ln2,且对任意的x,y∈R有f(x+y)=f(x)f(y),求f(x)
来源:学生作业帮 编辑:搜搜考试网作业帮 分类:数学作业 时间:2024/07/21 14:23:05
f(x)在x=0处可导且f'(0)=ln2,且对任意的x,y∈R有f(x+y)=f(x)f(y),求f(x)
![f(x)在x=0处可导且f'(0)=ln2,且对任意的x,y∈R有f(x+y)=f(x)f(y),求f(x)](/uploads/image/z/18734386-58-6.jpg?t=f%28x%29%E5%9C%A8x%3D0%E5%A4%84%E5%8F%AF%E5%AF%BC%E4%B8%94f%27%280%29%3Dln2%2C%E4%B8%94%E5%AF%B9%E4%BB%BB%E6%84%8F%E7%9A%84x%2Cy%E2%88%88R%E6%9C%89f%28x%2By%29%3Df%28x%29f%28y%29%2C%E6%B1%82f%28x%29)
f(0+0)=f(0)f(0),f(0)=1或f(0)=0
假如f(0)=0,{f(0+h)-f(0)}/h={f(0)f(h)-f(0)}/h=0,则有f′(0)=0≠ln2,所以f(0)≠0
f(x+h)-f(x)=f(X)f(h)-f(x)=f(x)(f(h)-1)=f(x)(f(h)-f(0))
{f(x+h)-f(x)}/h=f(x)(f(h)-f(0))/h,h→0,有f′(x)=f(x)ln2, dy/dx=yln2,x=0,y=1
dy/y=ln2dx,积分得lny=xln2+c, x=0,y=1,c=0
lny=xln2,
y=2^x
假如f(0)=0,{f(0+h)-f(0)}/h={f(0)f(h)-f(0)}/h=0,则有f′(0)=0≠ln2,所以f(0)≠0
f(x+h)-f(x)=f(X)f(h)-f(x)=f(x)(f(h)-1)=f(x)(f(h)-f(0))
{f(x+h)-f(x)}/h=f(x)(f(h)-f(0))/h,h→0,有f′(x)=f(x)ln2, dy/dx=yln2,x=0,y=1
dy/y=ln2dx,积分得lny=xln2+c, x=0,y=1,c=0
lny=xln2,
y=2^x
定义在R上的函数f(x)对任意x,y∈R都有f(x+y)+f(x-y)=2f(x)*f(y),且f(0)≠0,判断f(x
f(x)是定义在R上的函数,对任意x,y∈R,f(x+y)+f(x-y)=2f(x)f(y)恒成立,且f(0)≠0求f(
定义在R上的函数f(x),对任意x,y ∈R有f(x+y)+f(x-y)=2f(x)*f(y)且f(0)不等于0,则f(
定义在R上的函数f(x)对任意x,y属于R都有f(x+y)+f(x-y)=2f(x)f(y),且f(0)≠0,判断f(x
定义在R上的函数f(x),满足对任意x y∈R恒有f(xy)=f(x)+f(y) 且f(x)不恒为0 求f(1)和f(-
已知f(x)对任意x,y属于R,总有f(x)+f(y)=f(x+y)且x>0时,f(x)
定义在R上的函数f(x),对任意的x、y∈R,有f(x+y)+f(x-y)=2f(x)f(y),且f(0)不等于0,求证
定义在R上的函数f(x),对任意的x、y∈R,有f(x+y)+f(x-y)=2f(x)f(y),且f(0)不等于1,求证
已知定义域在R上的函数f(x)对任意实数x,y,恒有f(x+y)+f(x-y)=2f(x)f(y),且f(0)不等于0求
设f (x )定义在R上的函数,且对任意x,y∈R,恒有f(x+y)=f(x)f(y),且x>0时,f(x)>1证明:
定义在R上的函数f(x) 对任意的x,y 且 属于R 有f(x+y)+f(x-y)=2f(x)f(y) 且f(0)不等于
已知定义在R上的单调函数y=f(x),当x1,且对任意的实数x,y∈R,有f(x+y)=f(x)f(y) 求f(0)