Question

dada a função de f: R em R definida por f(x)= -3x+1, calcule f(-1/3) e f(1/9)​

Answer 1

f(x)=-3x+1

Para o f(-1/3):

f(-1/3)= -3(-1/3)+1

f(-1/3)=1+1

f(-1/3)=2

Para o f(1/9):

f(1/9)= -3(1/9)+1

f(1/9)= -3/9+1

f(1/9)= -1/3+1

f(1/9)= -1/3+3/3

f(1/9)= 2/3

Bons estudos !

Answer 2

$f \: (x) = - 3x + 1 \\ f \: ( - \frac{1}{3} ) = - 3 \times ( - \frac{1}{3} ) + 1 \\ f \: ( - \frac{1}{3} ) = - \frac{3}{1} \times ( - \frac{1}{3} ) + 1 \\ f \: ( - \frac{1}{3} ) = \frac{3}{3} + 1 \\ f \: ( - \frac{1}{3} ) = 1 + 1 \\ f \: ( - \frac{1}{3} ) = 2$

$f \: (x) = - 3x + 1 \\ f \: ( \frac{1}{9} ) = - 3 \times ( \frac{1}{9} ) + 1 \\ f \: ( \frac{1}{9} ) = - \frac{3}{1 } \times ( \frac{1}{9} ) + 1 \\ f \: ( \frac{1}{9} ) = - \frac{3 \div 3}{9 \div 3} + 1 \\ f \: ( \frac{1}{9} ) = - \frac{1}{3} + 1 \\ f \: ( \frac{1}{9} ) = - \frac{1}{3} + \frac{1}{1} \\ f \: ( \frac{1}{9} ) = \frac{ - 1 + 3}{3} \\ f \: ( \frac{1}{9} ) = \frac{2}{3}$