Question

Alguém pode dar uma olhadinha nessa questão?

630c339b748878cc8ac9f682112ced11.docx

ad626903652e98d8bb3ec3a8c543cf17.docx

7fb8b5fed05ac160c922fa1c0e2c7375.docx

Answer 1

É dada a seguinte função:

$f(x)=|x-1|~~\Leftrightarrow~~f(x)=\left\{\! \begin{array}{lc} x-1\,,&\text{se }x\ge 1\\ 1-x\,,&\text{se }x< 1\\ \end{array} \right.$
__________________

Calculando as derivadas laterais:

$\bullet~~$ Quando $h\to 0$ pela direita, isto é

$h>0$

temos


$f'_+(1)=\underset{h\to 0^+}{\mathrm{\ell im}}~\dfrac{f(1+h)-f(1)}{h}\\\\\\ =\underset{h\to 0^+}{\mathrm{\ell im}}~\dfrac{|(1+h)-1|-|1-1|}{h}\\\\\\ =\underset{h\to 0^+}{\mathrm{\ell im}}~\dfrac{|h|-|0|}{h}\\\\\\ =\underset{h\to 0^+}{\mathrm{\ell im}}~\dfrac{|h|}{h}~~~~~(|h|=h\text{, pois }h>0)\\\\\\ =\underset{h\to 0^+}{\mathrm{\ell im}}~\dfrac{h}{h}\\\\\\ =\underset{h\to 0^+}{\mathrm{\ell im}}~1\\\\\\ \therefore~~f'_+(1)=1~~~~~~\mathbf{(i)}$

$\bullet~~$ Quando $h\to 0$ pela direita, isto é

$h<0$

temos


$f'_-(1)=\underset{h\to 0^-}{\mathrm{\ell im}}~\dfrac{f(1+h)-f(1)}{h}\\\\\\ =\underset{h\to 0^-}{\mathrm{\ell im}}~\dfrac{|(1+h)-1|-|1-1|}{h}\\\\\\ =\underset{h\to 0^-}{\mathrm{\ell im}}~\dfrac{|h|-|0|}{h}\\\\\\ =\underset{h\to 0^-}{\mathrm{\ell im}}~\dfrac{|h|}{h}~~~~~(|h|=-h\text{, pois }h<0)\\\\\\ =\underset{h\to 0^-}{\mathrm{\ell im}}~\dfrac{-h}{h}\\\\\\ =\underset{h\to 0^-}{\mathrm{\ell im}}~(-1)\\\\\\ \therefore~~f'_-(1)=-1~~~~~~\mathbf{(ii)}$
__________________


Por $\mathbf{(i)}$ e $\mathbf{(ii)},$ concluímos que

$f'(1)=\underset{h\to 0}{\mathrm{\ell im}}~\dfrac{f(1+h)-f(1)}{h}$

não existe (os limites laterais diferem).


Logo, $f$ não é derivável em $1.$