Question

Seja f(x)=1/x, x ≠ 0. Se f(2+p)-f(2)=3/2, então f(1-p)-f(1+p) é igual a?

Answer 1

Temos uma função, definida pela lei

$\mathsf{f(x)=\dfrac{1}{x}\qquad com~x\ne 0}$

Então, temos que

$\mathsf{f(2)=\dfrac{1}{2}}\\\\\\ \mathsf{f(2+p)=\dfrac{1}{2+p}}$

é só substituir o x pelo valor correspondente na função.

Sendo assim,

$\mathsf{f(2+p)-f(2)=\dfrac{3}{2}}\\\\\\ \mathsf{\dfrac{1}{2+p}-\dfrac{1}{2}=\dfrac{3}{2}}\\\\\\ \mathsf{\dfrac{1}{2+p}=\dfrac{3}{2}+\dfrac{1}{2}}\\\\\\ \mathsf{\dfrac{1}{2+p}=\dfrac{4}{2}}\\\\\\ \mathsf{\dfrac{1}{2+p}=2}$

$\mathsf{2(2+p)=1}\\\\ \mathsf{4+2p=1}\\\\ \mathsf{2p=1-4}\\\\ \mathsf{2p=-3}\\\\ \mathsf{p=-\,\dfrac{3}{2}}$

Então,

$\mathsf{f(1-p)-f(1+p)}\\\\\\ \mathsf{=f\bigg[1-\Big(-\dfrac{3}{2}\Big)\bigg]-f\bigg[1+\Big(-\dfrac{3}{2}\Big)\bigg]}\\\\\\ =\mathsf{f\Big(1+\dfrac{3}{2}\Big)-f\Big(1-\dfrac{3}{2}\Big)}\\\\\\ \mathsf{=f\Big(\dfrac{2}{2}+\dfrac{3}{2}\Big)-f\Big(\dfrac{2}{2}-\dfrac{3}{2}\Big)}\\\\\\ \mathsf{=f\Big(\dfrac{5}{2}\Big)-f\Big(\!-\!\dfrac{1}{2}\Big)}\\\\\\ \mathsf{=\dfrac{1}{~\frac{5}{2}~}-\dfrac{1}{-\frac{1}{2}}}\\\\\\ \mathsf{=\dfrac{2}{5}+2}\\\\\\ \mathsf{=\dfrac{2}{5}+\dfrac{10}{5}}$

$\mathsf{=\dfrac{12}{5}\quad\longleftarrow\quad resposta.}$

Dúvidas? Comente.

Bons estudos! :-)