mean value theorem and limit

Tony

2011-11-22 10:48 pm

suppose y=f(x) has continuous first and second order derivatives on
the interval (-1,1) and second order derivative (i.e.f''(x))is not equal to 0.
(a)show,for all nonzero x on the interval (-1,1),
there exists a unique z=z(x) on the interval (0,1)
such that f(x)=f(0)+xf'(zx)
where f'(zx) is the first order derivative of f(zx).
(b)show lim(x->0)z(x)=1/2

回答 (1)

吱

2011-11-23 10:07 pm

✔ 最佳答案

希望能幫到你........

圖片參考：http://latex.codecogs.com/gif.latex?0<z<1$ and $0<|x|<1\\$Then $0<zx<x<1$ or $-1<x<zx<0\\\\ $When $0<zx<x$ , $\frac{f(x)-f(0)}{x-0}=f'(zx)\Rightarrow f(x)=f(0)+xf'(zx)\\\\ $When $x<zx<0$ , $\frac{f(0)-f(x)}{0-x}=f'(zx)\Rightarrow f(x)=f(0)+xf'(zx)\\\\ (b)$ Differentiate w.r.t.$x$ , $f'(x)=f'(zx)+(z+z'x)xf''(zx)\\\\ f'(x)-f'(0)=f'(zx)-f'(0)+(z+z'x)xf''(zx)\\\\ \frac{f'(x)-f'(0)}{x}=z\frac{f'(zx)-f'(0)}{zx}+(z+z'x)f''(zx)\\\\ f''(0)=zf''(0)+zf''(0)$ , where $x\to0\\\\ \because f''(0)\neq0$ , $\therefore z=\frac{1}{2}

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