Question

gente alguém poderia me ajudar com a questão em anexo?

Answer 1

Boa noite!


Solução!


$f(x)=2x-1\\\\\\\ \dfrac{f(x+n)-f(x)}{n}\\\\\\ \dfrac{2(x+n)-1-2x+1)}{n}\\\\\\ \dfrac{(2x-2x-1+1+2n)}{n}\\\\\\ \dfrac{2x-2x-1+1+2n}{n}\\\\\\\ \dfrac{0+0+2n}{n}\\\\\\ \dfrac{2n}{n}=2$


Boa noite!
Bons estudos!

Answer 2

Oi Alice

f(x) = 2x - 1

f(x + n) = 2*(x + n) - 1 = 2x + 2n - 1

f(x + n) - f(x) = 2x + 2n - 1 - 2x + 1 = 2n

(f(x + n) - f(x))/n = 2n/n = 2

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