作业帮请求解决高数, f(x)在负无穷到正无穷上连续,且f[f(x)]=x证明至少存在一点a属于负无穷到正无穷,使f(a)=a.f(x)在0到正无穷上有定义,且f ' (1)=a!=0,对任意x,y属于0到正无穷满足f(xy)=f(x)+f(y),求f(x).
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![请求解决高数, f(x)在负无穷到正无穷上连续,且f[f(x)]=x证明至少存在一点a属于负无穷到正无穷,使f(a)=a.f(x)在0到正无穷上有定义,且f ' (1)=a!=0,对任意x,y属于0到正无穷满足f(xy)=f(x)+f(y),求f(x).](/uploads/image/z/1823223-39-3.jpg?t=%E8%AF%B7%E6%B1%82%E8%A7%A3%E5%86%B3%E9%AB%98%E6%95%B0%2C+++f%28x%29%E5%9C%A8%E8%B4%9F%E6%97%A0%E7%A9%B7%E5%88%B0%E6%AD%A3%E6%97%A0%E7%A9%B7%E4%B8%8A%E8%BF%9E%E7%BB%AD%2C%E4%B8%94f%5Bf%28x%29%5D%3Dx%E8%AF%81%E6%98%8E%E8%87%B3%E5%B0%91%E5%AD%98%E5%9C%A8%E4%B8%80%E7%82%B9a%E5%B1%9E%E4%BA%8E%E8%B4%9F%E6%97%A0%E7%A9%B7%E5%88%B0%E6%AD%A3%E6%97%A0%E7%A9%B7%2C%E4%BD%BFf%28a%29%3Da.f%28x%29%E5%9C%A80%E5%88%B0%E6%AD%A3%E6%97%A0%E7%A9%B7%E4%B8%8A%E6%9C%89%E5%AE%9A%E4%B9%89%2C%E4%B8%94f+%26%2339%3B+%281%29%3Da%21%3D0%2C%E5%AF%B9%E4%BB%BB%E6%84%8Fx%2Cy%E5%B1%9E%E4%BA%8E0%E5%88%B0%E6%AD%A3%E6%97%A0%E7%A9%B7%E6%BB%A1%E8%B6%B3f%28xy%29%3Df%28x%29%2Bf%28y%29%2C%E6%B1%82f%28x%29.)
请求解决高数, f(x)在负无穷到正无穷上连续,且f[f(x)]=x证明至少存在一点a属于负无穷到正无穷,使f(a)=a.f(x)在0到正无穷上有定义,且f ' (1)=a!=0,对任意x,y属于0到正无穷满足f(xy)=f(x)+f(y),求f(x).
请求解决高数,
f(x)在负无穷到正无穷上连续,且f[f(x)]=x证明至少存在一点a属于负无穷到正无穷,使f(a)=a.
f(x)在0到正无穷上有定义,且f ' (1)=a!=0,对任意x,y属于0到正无穷满足f(xy)=f(x)+f(y),求f(x).
请求解决高数, f(x)在负无穷到正无穷上连续,且f[f(x)]=x证明至少存在一点a属于负无穷到正无穷,使f(a)=a.f(x)在0到正无穷上有定义,且f ' (1)=a!=0,对任意x,y属于0到正无穷满足f(xy)=f(x)+f(y),求f(x).
因为 f[f(x)] = x,所以 f[f(x)]-f(x) = x-f(x),任取一点 x0,若 f(x0) = x0,则已找到 a = x0,使 f(a) = a；否则设 f(x0) = x1,此时 x0 ≠ x1.于是 f(x1)-x1 = x0-f(x0),也即 f(x1)-x1 与 f(x0)-x0 符号相反且均不为 0.设 g(x) = f(x)-x,则 g(x) 在两点 x0,x1 处反号,于是存在一点 a 属于 (x0,x1),使 g(a) = 0,所以存在一点 a,使 f(a) = a.
取 e^x,e^y 代入 f(xy) = f(x)+f(y),于是有 f(e^(x+y)) = f(e^x)+f(e^y),设 g(x) = f(e^x),那么 g(x+y) = g(x)+g(y)；取 y = dx,那么 g(x+dx) = g(x)+g(dx) = g(x)+g'(0)dx,所以 g'(x) = g'(0),也即 g(x) = g'(0)x+C,所以 f(x) = g(ln(x)) = Aln(x)+C,代入 f'(1) = a,得到 A = a,从而 f(x) = aln(x)+C,又 f(1) = f(1)+f(1),知道 f(1) = 0,所以 C = 0,求得 f(x) = aln(x).