Question

dada a função f(x)=1/x-2+ 1/x+3
determine seu dominio ?
qual o valor de f(-1)?
determine o valor f(1)=f(0) /f(-1)-f(-2)

Answer 1

$Dominio    x-2=0   x=2 \\ \\ x+3=0 \ \ x=-3 \\ \\  x \neq 2,x \neq -3 \\ \\  D_{f} = x\ \textless \ -3 ou -3\ \textless \ x\ \textless \ 2 ou x\ \textgreater \ 2   \\ \\ f(-1) = \frac{1}{-1-2} + \frac{1}{-1+3} = \frac{-1}{3} + \frac{1}{2} = \frac{-2+3}{6}= \frac{1}{6} \\ \\ f(1) = \frac{1}{1-2} + \frac{1}{1+3}= \frac{1}{-1} + \frac{1}{4} = -1+ \frac{1}{4} = \frac{-4+1}{4} = - \frac{3}{4}  \\ \\ f(0) = \frac{1}{0-2} + \frac{1}{0+3} \\ \\ f(0) = - \frac{1}{2} + \frac{1}{3} = \frac{-3+2}{6} = - \frac{1}{6}$
$f(-2) = \frac{1}{-2-2} + \frac{1}{-2+3} \\ \\ f(-2) = - \frac{1}{4} +1= \frac{-1+4}{4} = \frac{3}{4} \\ \\ f(1) = \frac{f(0)}{f(-1)-f(-2)}$ 

A fórmula está correta o sinal de igual está certo

Answer 2

$f(x)=\frac{1}{x-2} + \frac{1}{x+3} \\ \\ x -2 \neq 0 \\ x \neq 2 \\ \\ x+3 \neq 0 \\ x \neq -3$

D=[x∈R | x ≠ 2, x ≠ - 3]

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$f(x)=\frac{1}{x-2} + \frac{1}{x+3} \\ \\ f(-1)=\frac{1}{-3} + \frac{1}{2} \\ \\ f(-1)=\frac{2 - 3}{-6} \\ \\ f(-1) = \frac{1}{6}$

$f(1)=\frac{1}{1-2} + \frac{1}{1+3} = -\frac{3}{4}$

$f(0)=\frac{1}{-2} + \frac{1}{3} = -\frac{1}{6}$

$f(-2)=\frac{1}{-4} + \frac{1}{1} = \frac{3}{4}$

$f(x) = \frac{f(0)}{f(-1)-f(-2)} = \frac{-\frac{1}{6}}{\frac{1}{6} - \frac{3}{4}} = \frac{ -\frac{1}{6} }{ \frac{-7}{12}} = \frac{-12}{-42} = \frac{2}{7}$