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Considere \[ f(x) = \begin{cases} x^2 &\text {se $x \leq 1$,}\\ 3x-2 &\text {se $x > 1$.} \end{cases} \]
Determine se \(f\) é derivável em \(x=4\) e, se for, encontre o valor de sua derivada nesse
ponto.
Se \(f\) for derivável, dê o valor abaixo. Caso contrário, escreva ”DNE”.
\(f'(4) = \class{xm-answer}{\cssId{xm-answer-1}{\phantom{\rule{7em}{1.5em}}}}\)
Considere \[ f(x) = \begin{cases} x^2 &\text {se $x \leq 1$,}\\ 3x-2 &\text {se $x > 1$.} \end{cases} \]
Determine se \(f\) é derivável em \(x=1\) e, se for, encontre o valor de sua derivada nesse
ponto.
Usando a definição por limite para encontrar a derivada de \(f\), obtemos \[ f'(1)= \lim _{x\to 1}\frac {f(\class{xm-answer}{\cssId{xm-answer-2}{\phantom{\rule{7em}{1.5em}}}})-1}{\class{xm-answer}{\cssId{xm-answer-3}{\phantom{\rule{7em}{1.5em}}}}}\\ \]
Não conseguimos encontrar uma única expressão para \(f(x)\), porque, quando \(x\) se
aproxima de \(1\), ele pode ser maior ou menor que \(1\), e \(f(x)\) tem expressões diferentes
para \(x\le 1\) e \(x>1\). Assim, temos de calcular os dois limites laterais separadamente.
\begin{align*} \lim _{x\to 1^{-}}\frac {f(\class{xm-answer}{\cssId{xm-answer-4}{\phantom{\rule{7em}{1.5em}}}})-1}{x-1}&= \lim _{x\to 1^{-}}\frac {(\class{xm-answer}{\cssId{xm-answer-5}{\phantom{\rule{7em}{1.5em}}}})-1}{x-1}\\ &= \lim _{x\to 1^{-}}\frac {(\class{xm-answer}{\cssId{xm-answer-6}{\phantom{\rule{7em}{1.5em}}}})(x-1)}{x-1}\\ &= \lim _{x\to 1^{-}}(x+1)\\ &=\class{xm-answer}{\cssId{xm-answer-7}{\phantom{\rule{7em}{1.5em}}}} \end{align*}
Por outro lado,
\begin{align*} \lim _{x\to 1^{+}}\frac {f(\class{xm-answer}{\cssId{xm-answer-8}{\phantom{\rule{7em}{1.5em}}}})-1}{x-1}&= \lim _{x\to 1^{+}}\frac {(\class{xm-answer}{\cssId{xm-answer-9}{\phantom{\rule{7em}{1.5em}}}})-1}{x-1}\\ &=\lim _{x\to 1^{+}}\frac {\class{xm-answer}{\cssId{xm-answer-10}{\phantom{\rule{7em}{1.5em}}}}}{x-1}\\ &=\lim _{x\to 1^{+}}\frac {3(\class{xm-answer}{\cssId{xm-answer-11}{\phantom{\rule{7em}{1.5em}}}})}{x-1}\\ &=\class{xm-answer}{\cssId{xm-answer-12}{\phantom{\rule{7em}{1.5em}}}} \end{align*}
Os dois limites laterais são diferentes e, portanto, \[ f'(1)= {\class{xm-answer}{\cssId{xm-answer-13}{\phantom{\rule{7em}{1.5em}}}}}\\ \]