Question

find The domain of the function.
f (x) =(x² - 2+3)/(x²-1)​

Answer 1

Hy mate here is your answer

Equate the denominator(x2−2x−3) to zero, then solve the equation for x

Equate the denominator(x2−2x−3) to zero, then solve the equation for x→x2−2x−3=0

Equate the denominator(x2−2x−3) to zero, then solve the equation for x→x2−2x−3=0→x=−(−2)±√(−2)2−4⋅(1)⋅(−3)2⋅1

Equate the denominator(x2−2x−3) to zero, then solve the equation for x→x2−2x−3=0→x=−(−2)±√(−2)2−4⋅(1)⋅(−3)2⋅1→x=1±2

Equate the denominator(x2−2x−3) to zero, then solve the equation for x→x2−2x−3=0→x=−(−2)±√(−2)2−4⋅(1)⋅(−3)2⋅1→x=1±2⇒x=−1 and x=3

Equate the denominator(x2−2x−3) to zero, then solve the equation for x→x2−2x−3=0→x=−(−2)±√(−2)2−4⋅(1)⋅(−3)2⋅1→x=1±2⇒x=−1 and x=3This means that, when x=−1 or 3, we have the x2−2x−3=0

Equate the denominator(x2−2x−3) to zero, then solve the equation for x→x2−2x−3=0→x=−(−2)±√(−2)2−4⋅(1)⋅(−3)2⋅1→x=1±2⇒x=−1 and x=3This means that, when x=−1 or 3, we have the x2−2x−3=0Implying that f(x)=x0 which is undefined.

Equate the denominator(x2−2x−3) to zero, then solve the equation for x→x2−2x−3=0→x=−(−2)±√(−2)2−4⋅(1)⋅(−3)2⋅1→x=1±2⇒x=−1 and x=3This means that, when x=−1 or 3, we have the x2−2x−3=0Implying that f(x)=x0 which is undefined.Hence, the domain is all real numbers except −1

Equate the denominator(x2−2x−3) to zero, then solve the equation for x→x2−2x−3=0→x=−(−2)±√(−2)2−4⋅(1)⋅(−3)2⋅1→x=1±2⇒x=−1 and x=3This means that, when x=−1 or 3, we have the x2−2x−3=0Implying that f(x)=x0 which is undefined.Hence, the domain is all real numbers except −1and 3

Equate the denominator(x2−2x−3) to zero, then solve the equation for x→x2−2x−3=0→x=−(−2)±√(−2)2−4⋅(1)⋅(−3)2⋅1→x=1±2⇒x=−1 and x=3This means that, when x=−1 or 3, we have the x2−2x−3=0Implying that f(x)=x0 which is undefined.Hence, the domain is all real numbers except −1and 3Also written as Df=(−∞,−1)∪(−1,3)∪(3,+∞)

Equate the denominator(x2−2x−3) to zero, then solve the equation for x→x2−2x−3=0→x=−(−2)±√(−2)2−4⋅(1)⋅(−3)2⋅1→x=1±2⇒x=−1 and x=3This means that, when x=−1 or 3, we have the x2−2x−3=0Implying that f(x)=x0 which is undefined.Hence, the domain is all real numbers except −1and 3Also written as Df=(−∞,−1)∪(−1,3)∪(3,+∞)To Find the Range

Equate the denominator(x2−2x−3) to zero, then solve the equation for x→x2−2x−3=0→x=−(−2)±√(−2)2−4⋅(1)⋅(−3)2⋅1→x=1±2⇒x=−1 and x=3This means that, when x=−1 or 3, we have the x2−2x−3=0Implying that f(x)=x0 which is undefined.Hence, the domain is all real numbers except −1and 3Also written as Df=(−∞,−1)∪(−1,3)∪(3,+∞)To Find the RangeStep 1

Equate the denominator(x2−2x−3) to zero, then solve the equation for x→x2−2x−3=0→x=−(−2)±√(−2)2−4⋅(1)⋅(−3)2⋅1→x=1±2⇒x=−1 and x=3This means that, when x=−1 or 3, we have the x2−2x−3=0Implying that f(x)=x0 which is undefined.Hence, the domain is all real numbers except −1and 3Also written as Df=(−∞,−1)∪(−1,3)∪(3,+∞)To Find the RangeStep 1say f(x)=y and rearrange the function as a quadratic equation

Equate the denominator(x2−2x−3) to zero, then solve the equation for x→x2−2x−3=0→x=−(−2)±√(−2)2−4⋅(1)⋅(−3)2⋅1→x=1±2⇒x=−1 and x=3This means that, when x=−1 or 3, we have the x2−2x−3=0Implying that f(x)=x0 which is undefined.Hence, the domain is all real numbers except −1and 3Also written as Df=(−∞,−1)∪(−1,3)∪(3,+∞)To Find the RangeStep 1say f(x)=y and rearrange the function as a quadratic equationy=xx2−2x−3

Equate the denominator(x2−2x−3) to zero, then solve the equation for x→x2−2x−3=0→x=−(−2)±√(−2)2−4⋅(1)⋅(−3)2⋅1→x=1±2⇒x=−1 and x=3This means that, when x=−1 or 3, we have the x2−2x−3=0Implying that f(x)=x0 which is undefined.Hence, the domain is all real numbers except −1and 3Also written as Df=(−∞,−1)∪(−1,3)∪(3,+∞)To Find the RangeStep 1say f(x)=y and rearrange the function as a quadratic equationy=xx2−2x−3→y(x2

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