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If $f\,'$ exists on $I= [a,b]$, then $(\inf f\,'(I),\sup f\,'(I))\subset f\,'(I)$
04-05-2017, 11:49 AM
Post: #1
If $f\,'$ exists on $I= [a,b]$, then $(\inf f\,'(I),\sup f\,'(I))\subset f\,'(I)$
$\quad$Show that if $f\,'$ exists on $I = [a,b]$, then
$\quad a,b\in f'(I)\implies [\min(a,b),\max(a,b)]\subset f'(I).$
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04-05-2017, 12:19 PM
Post: #2
The Proof:
Asume that $a,b\in I,\underset{\,}{\;} a< b,\; \xi\in(f\,'(a) f\,'(b))$, then
$\small\underset{\,}{\dfrac{f(a+h)-f(a)}{h}}< \xi < \dfrac{f(b-h)-f(b)}{-h}$ for some$\,h\,$with
$h\in(0,b-a).$ Let $F(t) = \small\dfrac{f(t+h)-f(t)}{h}\;(a\le t\le b-h),$
then $F$ is continuous and $F(a)<\xi < F(b-h)$ thus
$\xi=F(\eta)={\small\dfrac{f(\eta+h)-f(\eta)}{h}}=f\,'(\eta+\theta h)$ for some
$\eta\in(a,b-h),\;\theta\in (0,1)$ thus $\eta+\theta h\in (a,b).\quad\square$
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