Rational functions #5 - Questions and Answers

Question 1 prove that for all rational numbers r and s, rs is rational Answer Given that r and s are rational numbers. That is r and s can be written in the form r = a/b and s = c/d, b,d is not zero. Obtain the value of r x s as shown below. r x s = (a/b) x (c/d) r x s = ac/bd Note that, b, d is not equal to zero. Therefore, bd is not zero. Further, note that as a, b, c and d are integers ac and bd are also integers. Step 2 Therefore, note that the product r x s is the ratio of two integers with denominator not equal to zero. It is known that, the above is a definition of a rational number. Thus, the product r x s is rational. Hence proved. Question 2 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. Answer Let us take: f(x) = x / (x² + 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 Question 3 What are the Limits at Infinity of Rational Functions? Answer To Determine: What are the Limits at Infinity of Rational Functions? Step 2 Explanation: In case of rational function, given limit is infinity. For example: lim (x->∞) (4x² + 5) / (6x + 7) In that case, we will take common highest degree term (that is x²) to numerator and take common highest degree term (that is x) in the denominator. After that simplify it then take the limit. Question 4 A rational function is a function f of the form p/q where p and q are polynomial functions. The domain of f is {x ∈ R : q(x) ≠ 0}. Prove every rational function is continuous. I have previously proved that every polynomial function p(x) = a₀ + a₁x + ... + aₙxⁿ is continuous on R. Could I use a theorem that states f/g is continuous at z₀ if g(z₀) ≠ 0? Then p(x) (which is already proved to be continuous) over q(x) would be continuous? Answer A function f of x is continuous at x=a if and only if lim f(x) = f(a) x->a Obviously, as you've stated, not every rational function is continuous on the entire real line, so you just need to prove there is some interval I ⊂ R on which f is continuous-that is, there is an a ∈ I that satisfies the above criterion. So argue that there is at least one a ∈ R \ {x: q(x) = 0} such that lim p(x)/q(x) = p(a)/q(a) x->a lim p(x) / lim q(x) = p(a) / q(a) x->a x->a lim p(x) / lim q(x) ∈ R x->a x->a Since there is only a finite number of x: q(x) = 0 and since p(x) is defined across the whole real line, then there are an infinite number of valid a. Question 5 a) Show sin is not a rational function. By definition of a rational function, a rational function cannot be 0 at infinite points unless it is 0 everywhere. Obviously, sin have infinite points that are 0 and infinite points that are not zero, thus not a rational function. b) Show that there do not exist rational functions f₀, ..., fₙ-₁(x) (sinx)ⁿ⁻¹ + ... + f₀(x) = 0 for all x First choose x = 2kπ. So f₀(x) = 0 for x = 2kπ. Since f₀ is rational => f₀(x) = 0 for all x. Thus can write sin x [(sin x)ⁿ⁻¹ + fₙ-₁(sin x)ⁿ⁻² ... + f₁(x)] = 0 Question: The second factor is 0 for all x ≠ 2kπ. How does this imply that it is 0 for all x? And how does this lead to the result? Answer Use continuity of sin and rational functions.