Find the quotient function f/g for f(x)=sqr(x+1) and g(x)= sqr( x-1).
My Answer:
sqr(x+1)/sqr( x-1)
sqr(x^2-1)/ (x-1)
However, I also have to state the restrictions to the domain and range, which I do not know how to do. Could someone please help me? Thanks.
If Dom(f(x))=domain of f(x), and
Dom(g(x))=domain of g(x), the domain of f(x)/g(x) is the intersection
Dom(f(x))∩Dom(g(x))
However,by the definition of the quotient function, g(x) ≠ 0, therefore, we have to remove the members where g(x)=0 to get the final version:
Dom(f(x)/g(x))
= Dom(f(x))∩Dom(g(x)-{x:g(x)=0}
Note that:
Dom(√(x+1))
= [-1,∞]
Dom(√(x-1))
= [1,∞]
So
Dom(f(x)∩g(x))=[1,∞]
g(x)=0 when x=1, this has to be removed.
Therefore
Dom(f(x)/g(x))
=(1,∞)
To find the quotient function f/g for f(x) = √(x+1) and g(x) = √(x-1), you can use the quotient rule for radicals. The rule states that √a / √b = √(a/b), as long as b is not equal to 0.
Applying this rule, we can simplify the quotient function as follows:
f(x) / g(x) = (√(x+1)) / (√(x-1))
To remove the radicals from the denominator, we can multiply the numerator and denominator by the conjugate of the denominator, which in this case is (√(x-1)):
[f(x) / g(x)] * [(√(x-1)) / (√(x-1))] = [(√(x+1)) * (√(x-1))] / [(√(x-1)) * (√(x-1))]
Simplifying this expression further:
[f(x) / g(x)] * [(√(x-1)) / (√(x-1))] = (√[(x+1)*(x-1)]) / (√[(x-1)*(x-1)])
The numerator can be further simplified as √[(x+1)*(x-1)] = √(x^2 - 1).
Following the same simplification, the denominator becomes (√[(x-1)*(x-1)]) = √(x^2 - 1).
Therefore, the quotient function f/g simplifies to:
f(x) / g(x) = (√(x^2 - 1)) / (√(x^2 - 1))
This simplification indicates that the quotient function is equal to 1, except when x = 1. When x = 1, the denominator (√(x^2 - 1)) becomes 0, which would result in division by 0, making the quotient undefined.
Thus, the domain of the quotient function f/g is all real numbers except x = 1.