anonymous
  • anonymous
Find f(x) and g(x) so that the function can be described as y = f(g(x)). y = two divided by x squared + 9
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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schrodinger
  • schrodinger
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anonymous
  • anonymous
@Nnesha plssss
anonymous
  • anonymous
@agent0smith
Nnesha
  • Nnesha
\[y=\frac{ 2 }{ x^2 }+9\] like this ?

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More answers

anonymous
  • anonymous
yes
Nnesha
  • Nnesha
f(g(x)) meaning substitute x for g(x) function
anonymous
  • anonymous
yeah, and then you simplify?
Nnesha
  • Nnesha
right but this one is backwardz u have to find f(x) and g(x) from\[y=\frac{ 2 }{ x^2 }+9\]
anonymous
  • anonymous
is f(x) = 2/x + 9 and g(x) = x^2
Nnesha
  • Nnesha
hmm
Nnesha
  • Nnesha
could be this \[\frac{ 2 }{ x^2 } +9 \] g(x) =x hmmm
anonymous
  • anonymous
yeah do both work?
anonymous
  • anonymous
i got it right, it was whtai wrote :)
anonymous
  • anonymous
i have one more coudl you help wiht that?
Nnesha
  • Nnesha
hmm okay i'll try
anonymous
  • anonymous
ok thanks
anonymous
  • anonymous
Confirm that f and g are inverses by showing that f(g(x)) = x and g(f(x)) = x. f(x) = Quantity x plus four divided by six and g(x) = 6x - 4
anonymous
  • anonymous
f(x) = x+4 / 6 g(x) = 6x - 4
Nnesha
  • Nnesha
alright first try f(g(x))=x substitute x for g(x) function \[\huge\rm f(\color{ReD}{g(x}))=\frac{( \color{ReD}{6x-4} )+ 4}{ 6 }\] now solve right side is it equal to x ??
anonymous
  • anonymous
yes
Nnesha
  • Nnesha
alright now 2nd one g(f(x) ) is this equal to x ?
anonymous
  • anonymous
umm
anonymous
  • anonymous
yes?
Nnesha
  • Nnesha
i don't know it yet
Nnesha
  • Nnesha
substitute x for f(x) function into f(x)
anonymous
  • anonymous
ok
Nnesha
  • Nnesha
\[\huge\rm g(\color{ReD}{f(x)})=6\color{ReD}{x}-4\] replace x with (x+4)/6
anonymous
  • anonymous
6(x+4)/6 -4
Nnesha
  • Nnesha
right \[6(\frac{x+4 }{ 6 })-4\] is it equal to x ?
anonymous
  • anonymous
yers
Nnesha
  • Nnesha
both are equal that shows both are inverse of each other f(g(x)) = g(f(x))
anonymous
  • anonymous
thanks so muhc!
Nnesha
  • Nnesha
np :=)

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