Sarahisme

ok ok , i see what your saying and i agree, but lets say then i just want to solve the problem f(f(x)) = x is my two cases right? and is there more cases than that? thanks for you help matty

what do you mean? isnt the domain just values that do not make the denominator 0 in for f(x) and the codomain the range of those values in inverse of f(x)?? if not could please please plesae explain further...

ok ok, ignore all the above posts....my ones anyways ....and lets start fresh... so the question is: "For which values of the constants a, b and c is the function f(x) = (x − a)/(bx − c) self inverse?" so far all i can come up with is 2 cases: Case 1) c = 1 a,b = real numbers ab != 1 (!= means cannot equal, i.e. equals sign with a slash through it) and Case 2) c = -1 a = 0 b = 0 ab != 1

if anyone is around to help, i sorta need it now lol please...

how many cases are there? i get down to the formula b(x^2) + (c^2)x + a = bc(x^2) + x + ac and then i get the cases (i know there are more cases but these are the only ones i can find so far): Case 1: c = 1, a = any real number, b = any real number Case 2: c = 0, a = 0, b = 1/x and then i can't find anymore....but i am sure there must be some more

oops guess that didnt print out correctly, but yeah i think its sort of clear enough? lol maybe...

why should [math] f(f(x))=\frac{f(x)-a}{bf(x)-c}=\frac{\frac{x-a}{bx-c}-a}{b\frac{x-a}{bx-c}-c} = 1 [/math] why is that not equal to just x?

defintion?

thanks Douglas, nah don't worry bout the rest of it, i'll save you some hassle Thanks anyways though

right ok, yeah thanks, that makes sense now Cheers Spyman Sarah

lol, not to seem to thick, but i read through your answer to question 3, and i don't really get it completely..... damn my brain! "Would you expect this resistance to significantly affect the measurement of voltages in the above circuit?" is my answer to this quesiton right? its a few posts back , sorry for the hassling sarah

oh ok, thanks Spyman, sorry i didnt see it there before *embarrassed*, thanks again

hmm ok i see what your saying, well i suppose since its not a "real physical" problem, i.e. its a theoretical circuit so yeah ?

what i said was..... Since the resistance of the multimeter is so large, it would result in there being no voltage for the rest of the circuit beyond the multimeter.

3. When measuring voltages, the resistance of the multimeter is 10 MΩ on all ranges. Would you expect this resistance to significantly affect the measurement of voltages in the above circuit?

oops my bad, hang on....

actually i got b, what bout question 3???

is the answer to 2) by nothing?

anyone got any ideas for question 2 ???? ???

actually changed my mind, for part 1) i get 112ohm: current through: 4.01x10^-2 Amps voltage across: 4.49 Volts 450ohm: current through: 1.00x10^-2 Amps voltage across: 4.51 Volts 110ohm: current through: 3.01x10^-2 Amps voltage across: 3.31 Volts 40ohm: current through: 3.01x10^-2 Amps voltage across: 1.20 Volts hows that?!?!

ok i get total resistance of the system to be 222 ohms and then from that, the total current to be 9/222 = 4.05x10^-2 then from this work out that the voltage drop across the 112 ohm resistor is 4.54 V and the current through it is 4.05x10^-2 Amps and then the voltage through the 450 ohm resistor is 4.46 V and that the current through it is 9.91x10^-3 Amps and that the voltage through the 110 ohm resistor is 4.46V and the current through it is 4.05x10^-2 Amps and that the voltage through the 40ohm resistor is 4.46V and the current through it is

ok, how long will you be around for, i will try the questions now (using your above advice) and then if possible i could give some answers and you tell me if good or not? lol ??

for part 1) is the voltage across each resistor just 9V??

hey could i please some help with the following questions? thanks