Question

Find the domain and range of the function f(x) = x+1/ x−2

Answer 1

Answer:

When x=1, f(x) = 1+1/1-2

=-2

When x=2, f(x) =2+1/2-2

=0

Answer 2

Answer:

Given the function

$\sf \: f(x) = \frac{x + 1}{x - 2} \\ \bf \: this \: function \: will \: exist \: if \:and\:only \:if \\ \\ \sf \: x \in \: R \: \: \: \: and \\ \sf \: x - 2 \neq0 \\ \therefore \: \sf \: x \neq \: 2 \\ \red{ \boxed{\green{\sf so \: domain \: D_f \: = R - \{2 \}}}} \\ \\ \bf \: we \: assumed \: that \\ \sf \: y = f(x) = \frac{x + 1}{x - 2} \\ \therefore \: \sf \: y = \frac{x + 1}{x - 2} \\ \sf \implies \: y(x - 2) = x + 1 \\ \sf \implies \:xy - 2y = x + 1 \\ \sf \implies \: xy - x = 2y + 1 \\ \sf \implies \: x(y - 1) = 2y + 1 \\ \sf \implies \: x = \frac{2y + 1}{y - 1} \: \: \: \: - - - - (1) \\ \\ \bf \: eqn.(1) \: will \: exist \: if\:and \:only\:if \\ \\ \sf \: y \in \: R \: \: \: and \: \\ \sf \: y - 1 \neq \: 0 \\ \therefore \: \sf \: y \neq \: 1 \\ \\ \green{ \boxed{{\red { \sf \: so \: range \: R_f = R - \{1 \}}}}}$