Does the ordered triple (-8, 10, -3) satisfy the following system of equations?
1 answer:
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Consider the functions m and n:
i)
for x≠1.
the domain of m is all the real numbers except 1.
ii) n(x)=x-3 is a polynomial function, because x-3 is a linear polynomial.
The domain of any polynomial function is all real numbers. To determine the Range of this function, let c be a value in the range:
x-3=c, then x=3+c.
so if we want the function to produce c, we just set x=3+c, check:
n(3+c)=(3+c)-3=c.
This means that the Range of n is all real numbers.
What this means, is that the input (x) of the function n(x) can be any real number, and the output (n(x)) can also be any number.
so let n(x)=a, consider m(a):
m(a)=(a+5)/(a-1), clearly a≠1, this means n(x)=x-3≠1, that is x≠-4
This means that the domain of the composed function, m(n(x)) = say f(x),
is all Real numbers except 4.
Check: letting x be 4, would mean n(4)=4-3=1
which would mean m(n(4))=m(1)=(1+5)/(1-1)=6/0, which makes no sense.
Answer: Any function with domain R-{4}, has the same domain of (mon)(x)
i)
the domain of m is all the real numbers except 1.
ii) n(x)=x-3 is a polynomial function, because x-3 is a linear polynomial.
The domain of any polynomial function is all real numbers. To determine the Range of this function, let c be a value in the range:
x-3=c, then x=3+c.
so if we want the function to produce c, we just set x=3+c, check:
n(3+c)=(3+c)-3=c.
This means that the Range of n is all real numbers.
What this means, is that the input (x) of the function n(x) can be any real number, and the output (n(x)) can also be any number.
so let n(x)=a, consider m(a):
m(a)=(a+5)/(a-1), clearly a≠1, this means n(x)=x-3≠1, that is x≠-4
This means that the domain of the composed function, m(n(x)) = say f(x),
is all Real numbers except 4.
Check: letting x be 4, would mean n(4)=4-3=1
which would mean m(n(4))=m(1)=(1+5)/(1-1)=6/0, which makes no sense.
Answer: Any function with domain R-{4}, has the same domain of (mon)(x)