f(g(x)) type question
What did I do wrong?
you know that to verify that \(\large\color{slate}{ f^{-1}(x) }\) is indeed the inverse of \(\large\color{slate}{ f(x) }\), you would need to show that: \(\large\color{black}{ \color{red}{f(}\color{blue}{f^{-1}(x)}\color{red}{)} =\color{blue}{f^{-1}(\color{red}{f(x)})} }\)
Yes
The question is asking for me to show that f(g(x)) and g(f(x)) both equal x I think. Can I just show that they equal each other?
I think it would be the same exact thing.
I mean if each of them is equal to x, then they are equal to each other, but I think you actually need to show that each is equal to x.
The idea is to show that they reflect each other over the graph of x, saying over a line of y=x.
swagmaster47 is not asking if both are equal is asking if they're inverse of each other
and you'd do it as SolomonZelman said by getting of f(g(x)), you should end up with "x" and the other way around
on the 3rd line, your multiplication on the numerator above, went awry \(\bf f(g(x))=\cfrac{\frac{-3x-7}{x-1}-7} {\frac{-3x-7}{x-1}+3}\implies \cfrac{\frac{-3x-7+[(x-1)\cdot -7]}{\cancel{ x-1}}}{\frac{-3x-7+[3(x-1)]}{\cancel{ x-1}}} \\ \quad \\ \implies \cfrac{-3x-7-7x+7}{-3x-7+3x-3}\implies ?\)
-10x/10 then x
I mean -10x/-10
and then when I get to g(f(x)) = (-3( (x-7)/(x + 3) )-7)/( ( (x-7)/(x + 3) )-1) I multiply by (x+3)/(x+3) ?
would that look like (-3( (x-7) )-7(x+3) )/( (x-7) -(x+3) )
well... those (x-3) cancel out because is a division of fractions and \(\bf \cfrac{\frac{a}{b}}{\frac{c}{{\color{blue}{ d}}}}\implies \cfrac{a}{b}\cdot \cfrac{{\color{blue}{ d}}}{c}\) would cancel each otehr out in the multiplication
hmm x+3 I meant.. anyhow, so they cancel each other out
Oh, so it looks like this? (-3( (x-7) - 7)/( x - 7 ) - 1)
I would just remove b and d in your diagram right?
If b = d
yes, because in this case b = d yes
Ok I think I get it! thanks.
yw
Wait what happens to the -3 coeffecient?
Nvm, got the right answer
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