Question

URGENTE: determinar o valor de a que torna f(x) continua.

Answer 1

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$\Large\boxed{\sf{\underline{continuidade~de~uma~func_{\!\!,}\tilde ao}}}\\\boxed{\begin{array}{c}\sf Dizemos~que~uma~func_{\!\!,}\tilde ao~f(x)\\\sf\acute e~cont\acute inua~em~x=a~quando:\\\checkmark\sf f(a)~existe\\\checkmark\displaystyle\sf\lim_{x \to a}f(x)~existe~e\\\checkmark\displaystyle\sf\lim_{x \to a}f(x)=f(a)\end{array}}$

$\sf f(x)=\begin{cases}\sf\dfrac{\sqrt{x}-2}{x-4};~se~x>4\\\sf3x+a;~se~x\leq 4\end{cases}$

$\sf f(4)=3\cdot4+a=12+a\\\displaystyle\sf\lim_{x \to 4^+}f(x)=\lim_{x \to 4}\dfrac{\sqrt{x}-2}{x-4}\\\displaystyle\lim_{x \to 4}\dfrac{\diagup\!\!\!\!\!\!(\sqrt{x}-\diagup\!\!\!\!2)}{\diagup\!\!\!\!\!\!(\sqrt{x}-\diagup\!\!\!\!2)(\sqrt{x}+2)}=\dfrac{1}{\sqrt{4}+2}}=\dfrac{1}{2+2}=\dfrac{1}{4}$

$\displaystyle\sf\lim_{x \to 4^-}f(x)=\lim_{x \to 4}3x+a=12+a\\\sf para~ser~cont\acute inua~\lim_{x \to 4}f(x)~precisa~existir~portanto\\\displaystyle\lim_{x \to 4^-}f(x)=\lim_{x \to 4^+}f(x)\\\sf 12+a=\dfrac{1}{4}\\\sf 48+4a=1\\\sf 4a=1-48\\\sf 4a=-47\\\huge\boxed{\boxed{\boxed{\boxed{\sf a=-\dfrac{47}{4}\checkmark}}}}$