mathmath333
  • mathmath333
Functions
Mathematics
  • Stacey Warren - Expert brainly.com
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chestercat
  • chestercat
I got my questions answered at brainly.com in under 10 minutes. Go to brainly.com now for free help!
mathmath333
  • mathmath333
\(\large \color{black}{\begin{align} & \normalsize \text{if}\ f(x)=\dfrac{1}{g(x)}\ \text{then which of the following is correct} \hspace{.33em}\\~\\ &a.)\ f(f(g(f(x)))) =f(g(g(f(f(x)))))\hspace{.33em}\\~\\ &b.)\ f(g(g(f(f(x)))))=f(f(g(g(g(x))))) \hspace{.33em}\\~\\ &c.)\ g(g(f(f(g(f(x))))))=f(f(g(g(f(g(x)))))) \hspace{.33em}\\~\\ &d.)\ f(g(g(g(f(x)))))= g(g(f(f(f(x)))))\hspace{.33em}\\~\\ \end{align}}\)
dan815
  • dan815
lol really
mathmath333
  • mathmath333
my mind blasted out while typing the options.

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dan815
  • dan815
i want to say C
mathmath333
  • mathmath333
yea option c.) is right
mathmath333
  • mathmath333
how did u figured so fast.
dan815
  • dan815
okay lets see why the same number of gs and fs makes it equal
dan815
  • dan815
lol just thought itd be nice if the only one with same gs and fs on both sides was the right one
dan815
  • dan815
same number* of
mathmath333
  • mathmath333
lol , funny judgment instinct
mathmath333
  • mathmath333
solomon's reaction after seeing this question. |dw:1435247455337:dw|
dan815
  • dan815
can we work out somethign simpler
dan815
  • dan815
f(g(x)) = g(f(x))?
dan815
  • dan815
is that true or what is the relationship between these 2
anonymous
  • anonymous
You 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
  • mathmath333
i was thinking we could assume some values for this function
mathmath333
  • mathmath333
ya that one 1/x ,x.
dan815
  • dan815
oaky to clean it up a little the x is making it very annoying
dan815
  • dan815
|dw:1435248000948:dw|
dan815
  • dan815
g o (f o g) = (g o f) o g
dan815
  • dan815
do you know if this is true?
mathmath333
  • mathmath333
that looks liitle relief to eyes
mathmath333
  • mathmath333
yes it looks true
dan815
  • dan815
are you sure, that looks too convenient
mathmath333
  • mathmath333
yea i mean this is true [g o (f o g) = (g o f) o g] ,isn't it ?
dan815
  • dan815
are you saying that is true in general or its true for this function
mathmath333
  • mathmath333
true for this function
dan815
  • dan815
oh interesting!! how did u get that
mathmath333
  • mathmath333
i put g(x)=x,f(x)=1/x
dan815
  • dan815
thats cheating lol
dan815
  • dan815
@ganeshie8
mathmath333
  • mathmath333
but that type of cheating is allowed
dan815
  • dan815
hey i got the lat part done i think
mathmath333
  • mathmath333
u sure this is true ? f o g = f o f
mathmath333
  • mathmath333
\(\large \color{black}{\begin{align}& g(x)=\dfrac{1}{x}\hspace{.33em}\\~\\ & f(x)=x \hspace{.33em}\\~\\ & f(g(x))=f(\dfrac{1}{x})=\dfrac{1}{x} \hspace{.33em}\\~\\ & f(f(x))=f(x)=x \hspace{.33em}\\~\\ & f(g(x))\neq f(f(x)) \hspace{.33em}\\~\\ \end{align}}\)
dan815
  • dan815
i found my mistake plz ignore!
ganeshie8
  • ganeshie8
to me, option \(c\) makes sense only if \(f\) and \(g\) are "inverse functions" and not reciprocals of each other
ganeshie8
  • ganeshie8
*not simply reciprocals of each other
ganeshie8
  • ganeshie8
then 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
  • mathmath333
i have check option c.) with substitution \(x\) and \(\dfrac{1}{x}\) it looks correct
dan815
  • dan815
is the answer C?
mathmath333
  • mathmath333
yes i told earlier
dan815
  • dan815
making sure :)
ganeshie8
  • ganeshie8
does textbook give some explanation ?
mathmath333
  • mathmath333
yea wait
mathmath333
  • mathmath333
\(\large \color{black}{\begin{align} & \normalsize \text{Book: the number of g's and f's should be equal on the LHS and RHS} \hspace{.33em}\\~\\ & \normalsize \text{since both these functions are essentially inverse of each other} \hspace{.33em}\\~\\ \end{align}}\)
mathmath333
  • mathmath333
i wonder if this is some kind of property
ParthKohli
  • ParthKohli
Multiplicative inverse is definitely not functional inverse. Try to convert all g's into 1/f's and you'll see...
dan815
  • dan815
i wonder what it means essentially
ParthKohli
  • ParthKohli
It means that the author doesn't know how to do it either.
dan815
  • dan815
im wondering if we can show something weaker like f o g o f being 1/g o f o g for multiplicative inverses
dan815
  • dan815
not only is choice C just the same number of f and g but its symmetric too f and g rev so
dan815
  • dan815
complete 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
  • dan815
try picking some functions and checking it out, something that is not inverse
ganeshie8
  • ganeshie8
f(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
  • ParthKohli
\[f(x) = x+1\]\[g(x) = \frac{1}{x+1}\]\[f\circ g \circ f = f\left(\frac{1}{x+2}\right)= \frac{1}{x+2}+1\]And\[g\circ f \circ g = g\left(\frac{x+2 }{x+1}\right)\]So nah.
ParthKohli
  • ParthKohli
Oh haha.
mathmath333
  • mathmath333
is the question wrong
ParthKohli
  • ParthKohli
Definitely.
ganeshie8
  • ganeshie8
would the question really make sense if we replace "reciprocal" with "functional inverse" ?
ParthKohli
  • ParthKohli
Yeah. I hope the guy who copied the question didn't make the obvious mistake: using \(\cdots^{-1}\) and \(1/\cdots\) interchangeably.
ganeshie8
  • ganeshie8
\(\large \color{black}{\begin{align} & \normalsize \text{if}\ f(x) \text{ and g(x)} \ \text{ are inverse functions, then which of the following is correct} \hspace{.33em}\\~\\ &a.)\ f(f(g(f(x)))) =f(g(g(f(f(x)))))\hspace{.33em}\\~\\ &b.)\ f(g(g(f(f(x)))))=f(f(g(g(g(x))))) \hspace{.33em}\\~\\ &c.)\ g(g(f(f(g(f(x))))))=f(f(g(g(f(g(x)))))) \hspace{.33em}\\~\\ &d.)\ f(g(g(g(f(x)))))= g(g(f(f(f(x)))))\hspace{.33em}\\~\\ \end{align}}\)
ParthKohli
  • ParthKohli
Here 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.
ganeshie8
  • ganeshie8
that looks simple !
ParthKohli
  • ParthKohli
Yeah, so the verdict is out: the guy who framed the question didn't think about it.
anonymous
  • anonymous
I love functions

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