urgente função exponencial.... Pergunta de ideia dewonderland97

March 2022 1 27 Report

urgente função exponencial

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Verkylen $a) f(x)= (\frac{1}{2})^x\\\\f(-3)= (\frac{1}{2})^{-3}-->f(-3)=2^3-->f(-3)=8 \\\\ f(-2)= (\frac{1}{2})^{-2}-->f(-2)=2^2-->f(-2)=4\\\\f(-1)= (\frac{1}{2})^{-1}-->f(-1)=2^1-->f(-1)=2\\\\f(0)= (\frac{1}{2})^0-->f(0)=1\\\\f(1)= (\frac{1}{2})^1-->f(1)= \frac{1}{2}\\\\f(2)= (\frac{1}{2})^2-->f(2)=\frac{1}{4}\\\\f(3)= (\frac{1}{2})^3-->f(3)= \frac{1}{8}$

$b) f(x)=2^x\\\\f(-3)=2^{-3}-->f(-3)= \frac{1}{2}^3-->f(-3)= \frac{1}{8}\\\\f(-2)=2^{-2}-->f(-2)= \frac{1}{2}^2-->f(-2)= \frac{1}{4}\\\\f(-1)=2^{-1}-->f(-1)= \frac{1}{2}^1-->f(-1)= \frac{1}{2}\\\\f(0)=2^0-->f(0)=1\\\\f(1)=2^1-->f(1)=2\\\\f(2)=2^2-->f(2)=4\\\\
f(3)=2^3-->f(3)=8\\\\\\Renato.$

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me ajudemmmm ! please qualquer questao...

Pessoal me ajudem a saber se acertei , balanceamento

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