Question

URGENTE!! Seja a funçãoexponencial f: R ---> R, definida por f(x) = (1/3)^x , determine: a) f(2) b) f(-2) c) f (1/2)

Answer 1

Explicação passo-a-passo:

$f(x)=\left(\dfrac{1}{3}\right)^x$

a) $f(2)=\left(\dfrac{1}{3}\right)^2=\dfrac{1}{3^2}=\dfrac{1}{9}$

b) $f(-2)=\left(\dfrac{1}{3}\right)^{-2}=3^2=9$

c) $f\left(\dfrac{1}{2}\right)=\left(\dfrac{1}{3}\right)^{\frac{1}{2}}=\sqrt{\dfrac{1}{3}}=\dfrac{1}{\sqrt{3}}=\dfrac{\sqrt{3}}{3}$

Answer 2

⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀$f(x) ={\left ( \dfrac{1}{3}\right )}^{x}$

A) $f(2) = {\left(\dfrac{1}{3}\right )}^{2} = \dfrac{ {1}^{2} }{ {3}^{2} } = \dfrac{1}{9}$

B) $f( - 2) ={ \left( \dfrac{1}{3} \right)}^{ - 2} = \dfrac{{3}^{2} }{{1}^{2}}= \dfrac{9}{1} = 9$

C) $f( \frac{1}{2} ) ={\left ( \dfrac{1}{3}\right ) }^{ \frac{1}{2} } = \sqrt{ \left(\dfrac{1}{3} \right)} = \dfrac{1}{ \sqrt{3} } = \dfrac{ \sqrt{3}}{3}$

Espero Ter Ajudado !!