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mathmath333
 one year ago
Functions
mathmath333
 one year ago
Functions

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mathmath333
 one year ago
Best ResponseYou've already chosen the best response.1\(\large \color{black}{\begin{align} & \normalsize \text{if}\ f(x)=\dfrac{1}{g(x)}\ \text{then which of the following is correct} \hspace{.33em}\\~\\ &a.)\ f(f(g(f(x)))) =f(g(g(f(f(x)))))\hspace{.33em}\\~\\ &b.)\ f(g(g(f(f(x)))))=f(f(g(g(g(x))))) \hspace{.33em}\\~\\ &c.)\ g(g(f(f(g(f(x))))))=f(f(g(g(f(g(x)))))) \hspace{.33em}\\~\\ &d.)\ f(g(g(g(f(x)))))= g(g(f(f(f(x)))))\hspace{.33em}\\~\\ \end{align}}\)

mathmath333
 one year ago
Best ResponseYou've already chosen the best response.1my mind blasted out while typing the options.

mathmath333
 one year ago
Best ResponseYou've already chosen the best response.1yea option c.) is right

mathmath333
 one year ago
Best ResponseYou've already chosen the best response.1how did u figured so fast.

dan815
 one year ago
Best ResponseYou've already chosen the best response.4okay lets see why the same number of gs and fs makes it equal

dan815
 one year ago
Best ResponseYou've already chosen the best response.4lol just thought itd be nice if the only one with same gs and fs on both sides was the right one

mathmath333
 one year ago
Best ResponseYou've already chosen the best response.1lol , funny judgment instinct

mathmath333
 one year ago
Best ResponseYou've already chosen the best response.1solomon's reaction after seeing this question. dw:1435247455337:dw

dan815
 one year ago
Best ResponseYou've already chosen the best response.4can we work out somethign simpler

dan815
 one year ago
Best ResponseYou've already chosen the best response.4is that true or what is the relationship between these 2

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0You can pick for simplicity F(x)=x, g(x)=1/x.. I know it's not professional solution but th question who started being silly :P

mathmath333
 one year ago
Best ResponseYou've already chosen the best response.1i was thinking we could assume some values for this function

mathmath333
 one year ago
Best ResponseYou've already chosen the best response.1ya that one 1/x ,x.

dan815
 one year ago
Best ResponseYou've already chosen the best response.4oaky to clean it up a little the x is making it very annoying

dan815
 one year ago
Best ResponseYou've already chosen the best response.4g o (f o g) = (g o f) o g

dan815
 one year ago
Best ResponseYou've already chosen the best response.4do you know if this is true?

mathmath333
 one year ago
Best ResponseYou've already chosen the best response.1that looks liitle relief to eyes

dan815
 one year ago
Best ResponseYou've already chosen the best response.4are you sure, that looks too convenient

mathmath333
 one year ago
Best ResponseYou've already chosen the best response.1yea i mean this is true [g o (f o g) = (g o f) o g] ,isn't it ?

dan815
 one year ago
Best ResponseYou've already chosen the best response.4are you saying that is true in general or its true for this function

mathmath333
 one year ago
Best ResponseYou've already chosen the best response.1true for this function

dan815
 one year ago
Best ResponseYou've already chosen the best response.4oh interesting!! how did u get that

mathmath333
 one year ago
Best ResponseYou've already chosen the best response.1i put g(x)=x,f(x)=1/x

mathmath333
 one year ago
Best ResponseYou've already chosen the best response.1but that type of cheating is allowed

dan815
 one year ago
Best ResponseYou've already chosen the best response.4hey i got the lat part done i think

mathmath333
 one year ago
Best ResponseYou've already chosen the best response.1u sure this is true ? f o g = f o f

mathmath333
 one year ago
Best ResponseYou've already chosen the best response.1\(\large \color{black}{\begin{align}& g(x)=\dfrac{1}{x}\hspace{.33em}\\~\\ & f(x)=x \hspace{.33em}\\~\\ & f(g(x))=f(\dfrac{1}{x})=\dfrac{1}{x} \hspace{.33em}\\~\\ & f(f(x))=f(x)=x \hspace{.33em}\\~\\ & f(g(x))\neq f(f(x)) \hspace{.33em}\\~\\ \end{align}}\)

dan815
 one year ago
Best ResponseYou've already chosen the best response.4i found my mistake plz ignore!

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.2to me, option \(c\) makes sense only if \(f\) and \(g\) are "inverse functions" and not reciprocals of each other

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.2*not simply reciprocals of each other

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.2then we can say the "undo" and "redo" count match on each side but if they're just reciprocals of each other, i don't really see how option c is right

mathmath333
 one year ago
Best ResponseYou've already chosen the best response.1i have check option c.) with substitution \(x\) and \(\dfrac{1}{x}\) it looks correct

mathmath333
 one year ago
Best ResponseYou've already chosen the best response.1yes i told earlier

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.2does textbook give some explanation ?

mathmath333
 one year ago
Best ResponseYou've already chosen the best response.1\(\large \color{black}{\begin{align} & \normalsize \text{Book: the number of g's and f's should be equal on the LHS and RHS} \hspace{.33em}\\~\\ & \normalsize \text{since both these functions are essentially inverse of each other} \hspace{.33em}\\~\\ \end{align}}\)

mathmath333
 one year ago
Best ResponseYou've already chosen the best response.1i wonder if this is some kind of property

ParthKohli
 one year ago
Best ResponseYou've already chosen the best response.2Multiplicative inverse is definitely not functional inverse. Try to convert all g's into 1/f's and you'll see...

dan815
 one year ago
Best ResponseYou've already chosen the best response.4i wonder what it means essentially

ParthKohli
 one year ago
Best ResponseYou've already chosen the best response.2It means that the author doesn't know how to do it either.

dan815
 one year ago
Best ResponseYou've already chosen the best response.4im wondering if we can show something weaker like f o g o f being 1/g o f o g for multiplicative inverses

dan815
 one year ago
Best ResponseYou've already chosen the best response.4not only is choice C just the same number of f and g but its symmetric too f and g rev so

dan815
 one year ago
Best ResponseYou've already chosen the best response.4complete inverses will obey the rule no matter how the f and g are placed but maybe this relation will hold as long as we have this symmetry

dan815
 one year ago
Best ResponseYou've already chosen the best response.4try picking some functions and checking it out, something that is not inverse

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.2f(x) = x+1 g(x) = 1/(x+1) f(g(x)) = 1/(x+1)+1 = (x+2)/(x+1) g(f(x)) = 1/(x+1+1) = 1/(x+2) they don't match

ParthKohli
 one year ago
Best ResponseYou've already chosen the best response.2\[f(x) = x+1\]\[g(x) = \frac{1}{x+1}\]\[f\circ g \circ f = f\left(\frac{1}{x+2}\right)= \frac{1}{x+2}+1\]And\[g\circ f \circ g = g\left(\frac{x+2 }{x+1}\right)\]So nah.

mathmath333
 one year ago
Best ResponseYou've already chosen the best response.1is the question wrong

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.2would the question really make sense if we replace "reciprocal" with "functional inverse" ?

ParthKohli
 one year ago
Best ResponseYou've already chosen the best response.2Yeah. I hope the guy who copied the question didn't make the obvious mistake: using \(\cdots^{1}\) and \(1/\cdots\) interchangeably.

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.2\(\large \color{black}{\begin{align} & \normalsize \text{if}\ f(x) \text{ and g(x)} \ \text{ are inverse functions, then which of the following is correct} \hspace{.33em}\\~\\ &a.)\ f(f(g(f(x)))) =f(g(g(f(f(x)))))\hspace{.33em}\\~\\ &b.)\ f(g(g(f(f(x)))))=f(f(g(g(g(x))))) \hspace{.33em}\\~\\ &c.)\ g(g(f(f(g(f(x))))))=f(f(g(g(f(g(x)))))) \hspace{.33em}\\~\\ &d.)\ f(g(g(g(f(x)))))= g(g(f(f(f(x)))))\hspace{.33em}\\~\\ \end{align}}\)

ParthKohli
 one year ago
Best ResponseYou've already chosen the best response.2Here is what we do: whenever \(f(g(\cdots ))\) or \(g(f(\cdots)) \) occurs, we replace it with \(\cdots\). So (A): LHS \(f(f(x))\) RHS \(f(g(f(x)) = f(x)\) so nah. (B): LHS \(f(g(f(x)) = f(x)\) RHS \(f(g(g(x)) = g(x)\) so nah. (C): LHS \(x\) RHS \(x\) so yes.

ParthKohli
 one year ago
Best ResponseYou've already chosen the best response.2Yeah, so the verdict is out: the guy who framed the question didn't think about it.
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