Question

State the domain and range.
f(x)=5/x+3 +7

D = {x | x ≠ 3}; R = {y | y ≠ 5}

D = {x | x ≠ –3}; R = {y | y ≠ 5}

D = {x | x ≠ 5}; R = {y | y ≠ 7}

D = {x | x ≠ 3}; R = {y | y ≠ 7}

D = {x | x ≠ –3}; R = {y | y ≠ 7}

Answer 1

Interpreting f(x) as 5/(x+3) + 7:

Domain: denominator x+3 ≠ 0 ⇒ x ≠ −3, so D = {x | x ≠ −3}.

Range: y = 5/(x+3) + 7 ⇒ y − 7 = 5/(x+3). The left side cannot be 0 (5/(x+3) is never 0), so y ≠ 7. For any y ≠ 7 we can solve x = 5/(y−7) − 3, so every such y occurs. Thus R = {y | y ≠ 7}.

Answer: D = {x | x ≠ −3}; R = {y | y ≠ 7}.