Functions

- mathmath333

Functions

- jamiebookeater

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- 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

lol really

- mathmath333

my mind blasted out while typing the options.

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## More answers

- dan815

i want to say C

- mathmath333

yea option c.) is right

- mathmath333

how did u figured so fast.

- dan815

okay lets see why the same number of gs and fs makes it equal

- dan815

lol just thought itd be nice if the only one with same gs and fs on both sides was the right one

- dan815

same number* of

- mathmath333

lol , funny judgment instinct

- mathmath333

solomon's reaction after seeing this question.
|dw:1435247455337:dw|

- dan815

can we work out somethign simpler

- dan815

f(g(x)) = g(f(x))?

- dan815

is that true or what is the relationship between these 2

- 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

i was thinking we could assume some values for this function

- mathmath333

ya that one 1/x ,x.

- dan815

oaky to clean it up a little the x is making it very annoying

- dan815

|dw:1435248000948:dw|

- dan815

g o (f o g) = (g o f) o g

- dan815

do you know if this is true?

- mathmath333

that looks liitle relief to eyes

- mathmath333

yes it looks true

- dan815

are you sure, that looks too convenient

- mathmath333

yea i mean this is true [g o (f o g) = (g o f) o g] ,isn't it ?

- dan815

are you saying that is true in general or its true for this function

- mathmath333

true for this function

- dan815

oh interesting!! how did u get that

- mathmath333

i put g(x)=x,f(x)=1/x

- dan815

thats cheating lol

- dan815

- mathmath333

but that type of cheating is allowed

- dan815

hey i got the lat part done i think

- mathmath333

u sure this is true ?
f o g = f o f

- 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

i found my mistake plz ignore!

- ganeshie8

to me, option \(c\) makes sense only if \(f\) and \(g\) are "inverse functions" and not reciprocals of each other

- ganeshie8

*not simply reciprocals of each other

- 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

i have check option c.) with substitution \(x\) and \(\dfrac{1}{x}\) it looks correct

- dan815

is the answer C?

- mathmath333

yes i told earlier

- dan815

making sure :)

- ganeshie8

does textbook give some explanation ?

- mathmath333

yea wait

- 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

i wonder if this is some kind of property

- ParthKohli

Multiplicative inverse is definitely not functional inverse.
Try to convert all g's into 1/f's and you'll see...

- dan815

i wonder what it means essentially

- ParthKohli

It means that the author doesn't know how to do it either.

- 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

not only is choice C just the same number of f and g but its symmetric too f and g rev so

- 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

try picking some functions and checking it out, something that is not inverse

- 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

\[f(x) = x+1\]\[g(x) = \frac{1}{x+1}\]\[f\circ g \circ f = f\left(\frac{1}{x+2}\right)= \frac{1}{x+2}+1\]And\[g\circ f \circ g = g\left(\frac{x+2 }{x+1}\right)\]So nah.

- ParthKohli

Oh haha.

- mathmath333

is the question wrong

- ParthKohli

Definitely.

- ganeshie8

would the question really make sense if we replace "reciprocal" with "functional inverse" ?

- ParthKohli

Yeah. I hope the guy who copied the question didn't make the obvious mistake: using \(\cdots^{-1}\) and \(1/\cdots\) interchangeably.

- 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

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

that looks simple !

- ParthKohli

Yeah, so the verdict is out: the guy who framed the question didn't think about it.

- anonymous

I love functions

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