Question

Questão de função do 1°EM, alguem??????

Answer 1

a)

f(x) = $a^{x}$


9 = $a^{4}$

a = $\sqrt[4]{9}$


b)

f(-2) = $(\sqrt[4]{9})^{-2}$

f(-2) = $\frac{1}{(\sqrt[4]{9})^{2}}$

f(-2) = $\frac{1}{\sqrt[4]{(9)^{2}}}$

f(-2) = $\frac{1}{\sqrt[4]{81}}$

f(-2) = $\frac{1}{3}$



f(-1) = $(\sqrt[4]{9})^{-1}$

f(-1) = $\frac{1}{(\sqrt[4]{9})^{1}}$

f(-1) = $\frac{1}{\sqrt[4]{(9)^{1}}}$

f(-1) = $\frac{1}{\sqrt[4]{9}}$


Racionalizando:


f(-1) = $\frac{1}{\sqrt[4]{9}}$ . $\frac{\sqrt[4]{(9)^{3}}}{\sqrt[4]{(9)^{3}}}$

f(-1) = $\frac{\sqrt[4]{(9)^{3}}}{\sqrt[4]{(9)^{4}}}$

f(-1) = $\frac{\sqrt[4]{(3^{2})^{3}}}{9}$

f(-1) = $\frac{\sqrt[4]{(3)^{6}}}{9}$

f(-1) = $\frac{\sqrt[4]{(3)^{2} \ . \ (3)^{4}}}{9}$

f(-1) = $\frac{3 \ . \sqrt[4]{(3)^{2}}}{9}$

f(-1) = $\frac{3 \ . \sqrt[4]{9}}{9}$

f(-1) = $\frac{\sqrt[4]{9}}{3}$



f(0) = $(\sqrt[4]{9})^{0}$

f(0) = 1



f(1) = $(\sqrt[4]{9})^{1}$

f(1) = $\sqrt[4]{9}$



f(2) = $(\sqrt[4]{9})^{2}$

f(2) = $\sqrt[4]{(9)^{2}}$

f(2) = $\sqrt[4]{81}$

f(2) = 3