Construct a rational function whose domain is all real numbers. Provide evidence as to why the domain of your rational function is all real numbers.

Lewis Harvey 2021-03-06 Answered
Construct a rational function whose domain is all real numbers. Provide evidence as to why the domain of your rational function is all real numbers.
You can still ask an expert for help

Expert Community at Your Service

  • Live experts 24/7
  • Questions are typically answered in as fast as 30 minutes
  • Personalized clear answers
Learn more

Solve your problem for the price of one coffee

  • Available 24/7
  • Math expert for every subject
  • Pay only if we can solve it
Ask Question

Expert Answer

Aubree Mcintyre
Answered 2021-03-07 Author has 73 answers

Let us take: f(x)=xx2+1
For f(x) the domain is all real numbers because for this function denominator can never be 0.
If x2+1=0
then, x2=1
or, x=±(1)= no solution
That means there is no real number x for which x2+1=0.
So all real values are possible for this function. That gives domain= all real numbers

Not exactly what you’re looking for?
Ask My Question

Expert Community at Your Service

  • Live experts 24/7
  • Questions are typically answered in as fast as 30 minutes
  • Personalized clear answers
Learn more

Relevant Questions

asked 2021-02-25
True or False. The graph of a rational function may intersect a horizontal asymptote.
asked 2022-05-22
The given function is
f ( x , y ) = 1 x 2 + y 2 1
From my understanding, a rational function such as this one would not have any points in which f ( x , y ) = 0 , so I would conclude that no such points exist. Would this be true or am I not looking at this in a different way?
asked 2021-05-16
Construct a rational function that has a vertical asymptote at x = 3 and a removable discontinuity at x = -2. Explain how you determined your answer.
asked 2021-06-13

The process by which we determine limits of rational functions applies equally well to ratios containing noninteger or negative powers of x: Divide numerator and denominator by the highest power of x in the denominator and proceed from there. Find the limits.
limxx35x+32x+x234

asked 2022-05-15
Express the following rational function in continued-fraction form:
4 x 2 + 3 x 7 2 x 3 + x 2 x + 5
The answer is :
(inline continued fraction) 4 2 x 1 2 + 23 8 x 63 92 406 529 x + 33 23
which means
4 2 x 1 2 + 23 8 x 63 92 406 529 x + 33 23
asked 2022-05-13
This seems very obvious and I am having a bit of trouble producing a formal proof.
sketch proof that the composition of two polynomials is a polynomial
Let
p ( z 1 ) = a n z 1 n + a n 1 z 1 n 1 + . . . + a 1 z 1 + a 0 q ( z 2 ) = b n z 2 n + b n 1 z 2 n 1 + . . . + b 1 z 2 + b 0
be two complex polynomials of degree n where a n , . . , a 0 C and b n , . . , b o C .
Now,
( p q ) ( z 2 ) = p ( q ( z 2 ) )           (by definition) = a n ( q ( z 2 ) ) n + a n 1 ( q ( z 2 ) ) n 1 + . . . + a 1 ( q ( z 2 ) ) + a 0
which is clearly a complex polynomial of degree n 2 .
sketch proof that the composition of two rational functions is a rational function
A rational function is a quotient of polynomials.
Let
a ( z 1 ) = p ( z 1 ) q ( z 1 ) ,   b ( z 2 ) = p ( z 2 ) q ( z 2 )
Now,
( a b ) ( z 2 ) = a ( b ( z 2 ) )           (by definition) = p ( p ( z 2 ) q ( z 2 ) ) q ( p ( z 2 ) q ( z 2 ) ) = a n ( p ( z 2 ) q ( z 2 ) ) n + a n 1 ( p ( z 2 ) q ( z 2 ) ) n 1 + . . . + a 1 ( p ( z 2 ) q ( z 2 ) ) + a 0 b n ( p ( z 2 ) q ( z 2 ) ) n + b n 1 ( p ( z 2 ) q ( z 2 ) ) n 1 + . . . + b 1 ( p ( z 2 ) q ( z 2 ) ) + b 0
Notice that ( p ( z 2 ) q ( z 2 ) ) i         ( i = n , n 1 , . . , 0 ) is a polynomial as
( f g ) ( z 2 ) = f ( g ( z 2 ) ) = ( p ( z 2 ) q ( z 2 ) ) i
where
f ( x ) = x i ,     g ( z 2 ) = ( p ( z 2 ) q ( z 2 ) )
are both polynomials. Hence ( a b ) ( z 2 ) is a rational function as it is the quotient of polynomials.
asked 2022-02-16
I know that all meromorphic functions on CPP1 are rational functions. However, Im