Question

A função f: R* -> R é dada por f(x) = x + 1/x. Determine:
a) f(1/2)
b) f(x + 1), para qual x ≠ -1
c) f(a - 1), para qual a ≠ 1

Answer 1

$f(x)=x+\dfrac{1}{x}~~~~~(x\ne 0)$

a) $f\big(\frac{1}{2}\big):$

$f\left(\dfrac{1}{2}\right)=\dfrac{1}{2}+\dfrac{1}{\big(\frac{1}{2}\big)}\\\\\\ f\left(\dfrac{1}{2}\right)=\dfrac{1}{2}+2\\\\\\ f\left(\dfrac{1}{2}\right)=\dfrac{1}{2}+\dfrac{4}{2}\\\\\\ f\left(\dfrac{1}{2}\right)=\dfrac{1+4}{2}\\\\\\ \boxed{\begin{array}{c}f\left(\dfrac{1}{2}\right)=\dfrac{5}{2} \end{array}}$


b) $f(x+1):$

$f(x+1)=(x+1)+\dfrac{1}{x+1}\\\\\\ f(x+1)=\dfrac{(x+1)^2}{x+1}+\dfrac{1}{x+1}\\\\\\ f(x+1)=\dfrac{(x+1)^2+1}{x+1}\\\\\\ f(x+1)=\dfrac{x^2+2x+1+1}{x+1}\\\\\\ \boxed{\begin{array}{c} f(x+1)=\dfrac{x^2+2x+2}{x+1} \end{array}}~~~~~~(x\ne-1)$


c) $f(a-1):$

$f(a-1)=(a-1)+\dfrac{1}{a-1}\\\\\\ f(a-1)=\dfrac{(a-1)^2}{a-1}+\dfrac{1}{a-1}\\\\\\ f(a-1)=\dfrac{(a-1)^2+1}{a-1}\\\\\\ f(a-1)=\dfrac{a^2-2a+1+1}{a-1}\\\\\\ \boxed{\begin{array}{c}f(a-1)=\dfrac{a^2-2a+2}{a-1} \end{array}}~~~~~~(a\ne 1)$