Question

1) Dada a função afim f(x) = 2x + 2, determine: a) f(1)
b) f(0)
c) f(1/3)
d) f(-1/2)
help ​

Answer 1

Resposta:

$\boxed{\begin{cases}\mathtt{f(1) = 4}\\\mathtt{f(0) = 2}\\\\\mathtt{f\left(\dfrac{1}{3}\right) = \dfrac{8}{3}}\\\\\mathtt{f\left(-\dfrac{1}{2}\right) = 1}\\\end{cases}}$

Explicação passo a passo:

É nos dada a seguinte função:

$\mathtt{f(x) = 2x + 2}$

Então, para acharmos o valor de f(1), f(0), f(1/3) e f(-1/2), basta substituir na função dada pelo valor que está entre parênteses.

a) $\mathsf{f(x) = 2x + 2 \Rightarrow}\,\mathsf{f(1) = \,?}$

$\mathtt{f(x) = 2x + 2}\\\mathtt{f(1) = 2 \cdot 1 + 2}\\\mathtt{f(1) = 2 + 2}\\\\\boxed{\mathtt{f(1) = 4}}$

b) $\mathsf{f(x) = 2x + 2 \Rightarrow}\,\mathsf{f(0) = \,?}$

$\mathtt{f(x) = 2x + 2}\\\mathtt{f(0) = 2 \cdot 0 + 2}\\\mathtt{f(0) = 0 + 2}\\\\\boxed{\mathtt{f(0) = 2}}$

c) $\mathsf{f(x) = 2x + 2 \Rightarrow}\,\mathsf{f\left(\frac{1}{3}\right) = \,?}$

$\mathtt{f(x) = 2x + 2}\\\\\mathtt{f\left(\dfrac{1}{3}\right) = 2 \cdot \dfrac{1}{3} + 2}\\\\\mathtt{f\left(\dfrac{1}{3}\right) = \dfrac{2}{3} + 2}\\\\\mathtt{f\left(\dfrac{1}{3}\right) = \dfrac{2}{3} + \dfrac{6}{3}}\\\\\\\boxed{\mathtt{f\left(\dfrac{1}{3}\right) = \dfrac{8}{3}}}$

d) $\mathsf{f(x) = 2x + 2 \Rightarrow}\,\mathsf{f\left(-\frac{1}{2}\right) = \,?}$

$\mathtt{f(x) = 2x + 2}\\\\\mathtt{f\left(-\dfrac{1}{2}\right) = 2 \cdot \left(-\dfrac{1}{2}\right) + 2}\\\\\mathtt{f\left(-\dfrac{1}{2}\right) = -\dfrac{2}{2} + 2}\\\\\mathtt{f\left(-\dfrac{1}{2}\right) = -1 + 2}\\\\\\\boxed{\mathtt{f\left(-\dfrac{1}{2}\right) = 1}}$

Sendo assim, dada a função afim f(x) = 2x + 2, os valores de f(1), f(0), f(1/3) e f(-1/2) são, respectivamente, 4, 2, 8/3 e 1.

Dúvidas? Comente.