Question

sendo f (2x - 1 ) = 3x + 4, determine: a) f (x + 2 ) b) f (3x + 1 )

Answer 1

Vamos à resolução dos exercícios propostos.

Letra “a”

f(2x-1)=3x+4  =>

f[2(x/2+3/2)-1]=3(x/2+3/2)+4  =>

f[2(x+3/2)-1]=3(x+3/2)+4  =>

f[(x+3)-1]=3(x+3)/2+4  =>

f[x+2]=3(x+3)/2+4  =>

f(x+2)=3/2(x+3)+4  =>

f(x+2)=3x/2+9/2+8/2  =>

f(x+2)=3x/2+(9+8)/2  =>

f(x+2)=3x/2+17/2  

Letra “b”

f(2x-1)=3x+4  =>

f[2(3x/2+1)-1]=3(3x/2+1)+4  =>

f[3x+2-1]=9x/2+3+4  =>

f(3x+1)=9x/2+7  =>

Abraços!

Answer 2

Olá!

Seja

$g(x) = 2x - 1.$

Então temos que

$f(2x - 1) = f(g(x)) = 3x + 4$

Como

$g(x) = 2x - 1$

segue que

$2x - 1 = g(x) \\ 2x = g(x) + 1 \\ x = \frac{g(x) + 1}{2}$

e então

$f(g(x)) = 3 \times (\frac{g(x) + 1}{2} ) + 4 \\ f(g(x)) = \frac{3g(x) + 3}{2} + \frac{8}{2} \\ f(g(x)) = \frac{3g(x) + 11}{2}$


Ou seja,


$f(x) = \frac{3x + 11}{2}$

Agora podemos responder aos itens a) e b).

a)

$f(x + 2) = \frac{3(x + 2) + 11}{2} \\ f(x + 2) = \frac{3x + 6 + 11}{2} \\ f(x + 2) = \frac{3x + 17}{2}$


b)

$f(3x + 1) = \frac{3(3x + 1) + 11}{2} \\ f(3 x + 1) = \frac{9x + 3 + 11}{2} \\ f(3x + 1) = \frac{9x + 14}{2}$