Question

Find a rational function f: R ->R with range f(R) = [0,1]. (Thus f(x) = p(x)/q(x) for all x for suitable polynomials P and Q where Q has no real roots.

Answer 1

try using a square root somewhere in the answer.

Answer 2

or... an exponentiation!

Answer 3

i have literally no idea to start, can you help me start off first please?

Answer 4

http://lmgtfy.com/?q=Find+a+rational+function+f%3A+R+-%3ER+with+range+f(R)+%3D+%5B0%2C1%5D.+(Thus+f(x)+%3D+p(x)%2Fq(x)+for+all+x+for+suitable+polynomials+P+and+Q+where+Q+has+no+real+roots.+&l=1

Answer 5

hmm that one is broken. Try this one: http://lmgtfy.com/?q=Find+a+rational+function+f%3A+R+-%3ER+with+range+f(R)+%3D+%5B0%2C1%5D.+(Thus+f(x)+%3D+p(x)%2Fq(x)+for+all+x+for+suitable+polynomials+P+and+Q+where+Q+has+no+real+roots.+

Answer 6

thanks, I have to head to a lecture now but will look at it later!

Answer 7

Still doesn't work... :( oh well

let Q be defined as \[(\sqrt{x})^2 - 1\]

Answer 8

hmm that's still not right

Answer 9

\[Q(x) =\sqrt{1-x}\]

Answer 10

almost there

Answer 11

alright I got it :\[Q(x) = \sqrt{1-x^2}\]

Answer 12

P(X) can be anything,.

Answer 13

Why don't you take \(q(x)=x^2+a^2\), for any real number a?

Answer 14

\[f(x)=\frac{2x^2}{x^4+1}\]

Answer 15

Zarkon, could you have a look at the problem right below this one?

Answer 16

thanks guys. Q(x) = \[\sqrt{1-x ^{2}}\] and P can equal anything for example \[x^{3}\] ?