Question

Seja f: R → R definida por:

               x²- 4          se x ≥ 0
f(x)=
               - x              se x < 0

a f é derivavel no ponto x°=0? Provar

Answer 1

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$\large\begin{array}{l} \textsf{Dada a fun\c{c}\~ao }\mathsf{f:~}\mathbb{R}\to\mathbb{R}\end{array}$

$\large\begin{array}{l}\mathsf{f(x)=}\left\{\! \begin{array}{rl} \mathsf{x^2-4,}&\textsf{se }\mathsf{x\ge 0}\\ \mathsf{-x,}&\textsf{se }\mathsf{x\ \textless \ 0}\\ \end{array} \right. \end{array}$


$\large\begin{array}{l} \textsf{verificar se f \'e deriv\'avel em }\mathsf{x_0=0.}\\\\\\ \textsf{Para isso, o seguinte limite deve existir e ser finito:}\\\\ \mathsf{f'(x_0)=\underset{x\to x_0}{\ell im}~\dfrac{f(x)-f(x_0)}{x-x_0}}\\\\ \mathsf{f'(0)=\underset{x\to 0}{\ell im}~\dfrac{f(x)-f(0)}{x-0}\qquad\quad(i)} \end{array}$


•   $\large\textsf{Calculando o limite pela esquerda (por valores menores}$
$\large\begin{array}{l} \textsf{que 0):}\\\\ \mathsf{f_-'(0)=\underset{x\to 0^-}{\ell im}~\dfrac{f(x)-f(0)}{x-0}}\\\\ \mathsf{=\underset{x\to 0^-}{\ell im}~\dfrac{-x-(0^2-4)}{x}}\\\\ \mathsf{=\underset{x\to 0^-}{\ell im}~\dfrac{-x+4}{x}}\\\\ \mathsf{=\underset{x\to 0^-}{\ell im}~\dfrac{4-x}{x}}\\\\ \end{array}$

$\large\begin{array}{l} \mathsf{=\underset{x\to 0^-}{\ell im}~\dfrac{g(x)}{h(x)}\qquad\quad(ii)}\\\\\\ \textsf{sendo }\\\\\mathsf{g(x)=4-x~~e~~h(x)=x.} \end{array}$


$\large\begin{array}{l} \textsf{Temos uma indetermina\c{c}\~ao do tipo }\mathsf{\dfrac{k}{0},}\textsf{ com }\mathsf{k\ne 0.}\\\\ \textsf{Precisamos estudar o sinal desse quociente }\mathsf{\dfrac{g(x)}{h(x)}}\textsf{ na vizinhan\c{c}a}\\\textsf{\`a esquerda de }\mathsf{x_0=0.} \end{array}$


$\large\begin{array}{l} \textsf{Usando um quadro de sinais, temos que}\\\\ \begin{array}{cc} \mathsf{g(x)=4-x}\quad&\mathsf{\underline{~+++}\underset{0}{\circ}\underline{++++}\underset{4}{\bullet}\underline{---~\quad}_{\blacktriangleright}}\\\\ \mathsf{h(x)=x}\quad&\mathsf{\underline{~---}\underset{0}{\circ}\underline{++++}\underset{4}{\bullet}\underline{+++\quad}_{\blacktriangleright}}\\\\\\ \mathsf{\dfrac{g(x)}{h(x)}=\dfrac{4-x}{x}}\quad&\mathsf{\underline{~---}\underset{0}{\circ}\underline{++++}\underset{4}{\bullet}\underline{---\quad}_{\blacktriangleright}} \end{array} \end{array}$


$\large\begin{array}{l} \textsf{Notamos que o quociente \'e negativo \`a esquerda de}\\\mathsf{x_0=0.}\textsf{ Portanto, o limite (ii) \'e}\\\\ \mathsf{=\underset{x\to 0^-}{\ell im}~\dfrac{4-x}{x}=-\infty\not \in \mathbb{R}} \end{array}$


$\large\begin{array}{l} \textsf{Portanto, f n\~ao \'e deriv\'avel em }\mathsf{x_0=0.}\\\\\\ \textsf{D\'uvidas? Comente.}\\\\\\ \textsf{Bons estudos! :-)} \end{array}$


Tags:   função descontínua descontinuidade derivada não derivável limite quociente definição cálculo diferencial