Question

Sendo f (x) = 3x-1/ x^2+1, calcule :

a) f (2)
b) f ( -1)
c) f (1)+ f (3)
d) f (-1)+f (-2)

Answer 1

a)

$f(2) = \frac{3.2 - 1}{ {2}^{2} + 1 } = \frac{5}{5} = 1$

b)

$f( - 1) = \frac{3.( - 1) - 1}{ {( - 1)}^{2} + 1} = \frac{ - 4}{2} = - 2$

c)

$f(3) = \frac{3.3 - 1}{ {3}^{2} + 1} = \frac{8}{10} = \frac{4}{5}$

$f(1) = \frac{3.1 - 1}{ {1}^{2} + 1 } = \frac{2}{2} = 1$

$f(3) + f(1) = \frac{4}{5} + 1 = \frac{9}{5}$

d)

$f( - 2) = \frac{3.( - 2) - 1}{ {( - 2)}^{2} + 1 } = \frac{ - 7}{5}$

$f( - 1) + f( - 2) = - 2 - \frac{7}{5} = - \frac{17}{5}$

Answer 2

Resposta:

Explicação passo-a-passo:

$f(x) = \frac{3x - 1}{x^2 + 1} \\\\a) f(2)=\frac{3*(2) - 1}{(2^2) +1} = \frac{6-1}{4+1} = \frac{5}{5} = 1\\\\b) f(-1)=\frac{3*(-1) - 1}{(-1)^2 + 1} = \frac{-3 - 1}{1+1} = \frac{-4}{2} = -2\\\\f(1)+f(3) = (\frac{3*(1) -1}{(1^2)+1} ) + (\frac{3*(3)-1}{(3^2) +1}) = \frac{2}{2} +\frac{8}{10} = 1 + \frac{8}{10} = \frac{18}{10} = \frac{9}{5} = 1,8\\\\d) f(-1) + f(-2) = (\frac{3*(-1)-1}{(-1^2) + 1}) +(\frac{3*(-2) -1}{(-2^2)+1}) = (\frac{-4}{2}) + (\frac{-7}{5})= \frac{-20-14}{10} = \frac{-34}{10}= -\frac{17}{5} = -3,4$