Question

A função "f" é tal que $f(2x+3)=3x+2$. Nessas condições, $f(3x+2)$ é igual a : a) 3x+2 b) 2x+3 c) 2x/3 + 1 d) 9x/2 + 1/2

Answer 1

Chamemos 2x + 3 de k:

$2x+3=k\\2x=k-3\\x=(k-3)/2$

Logo:

$f(2x+3)=3x+2\\\\f(k)=3(\frac{k-3}{2})+2\\\\f(k)=\frac{3k-9}{2}+2~~~~~\therefore~~~~~\boxed{\boxed{f(x)=\dfrac{3x-9}{2}+2}}$
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Achando f(3x + 2):

$f(3x+2)=\dfrac{3(3x+2)-9}{2}+2\\\\\\f(3x+2)=\dfrac{9x+6-9}{2}+2\\\\\\f(3x+2)=\dfrac{9x-3}{2}+2\\\\\\f(3x+2)=\dfrac{9x}{2}-\dfrac{3}{2}+\dfrac{4}{2}\\\\\\f(3x+2)=\dfrac{9x}{2}+\dfrac{-3+4}{2}\\\\\\\boxed{\boxed{f(3x+2)=\dfrac{9x}{2}+\dfrac{1}{2}}}$

Answer 2

$f\left(2x+3\right )=3x+2\\ \\ f\left(2x+3\right )=\dfrac{3}{2}\cdot 2x+2\\ \\ f\left(2x+3\right )=\dfrac{3}{2}\cdot 2x+\dfrac{3}{2}\cdot 3-\dfrac{3}{2}\cdot 3+2\\ \\ f\left(2x+3 \right )=\dfrac{3}{2}\cdot\left(2x+3 \right )-\dfrac{9}{2}+2\\ \\ f\left(2x+3 \right )=\dfrac{3}{2}\cdot\left(2x+3 \right )+\dfrac{-9+4}{2}\\ \\ f\left(2x+3 \right )=\dfrac{3}{2}\cdot\left(2x+3 \right )-\dfrac{5}{2}$


Fazendo a substituição $u=2x+3$, chegamos a

$\boxed{f\left(u \right )=\dfrac{3}{2}u-\dfrac{5}{2}}$


Então,

$f\left(3x+2 \right )=\dfrac{3}{2} \cdot \left(3x+2 \right )-\dfrac{5}{2}\\ \\ f\left(3x+2 \right )=\dfrac{9x}{2}+\dfrac{6}{2}-\dfrac{5}{2}\\ \\ f\left(3x+2 \right )=\dfrac{9x}{2}+\dfrac{6-5}{2}\\ \\ \boxed{f\left(3x+2 \right )=\dfrac{9x}{2}+\dfrac{1}{2}}$


Resposta: alternativa $\text{d) }\dfrac{9x}{2}+\dfrac{1}{2}$.