Adi3
  • Adi3
will medal Please help. find the inverse of f(n) = 2(n-2)^3
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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jamiebookeater
  • jamiebookeater
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Adi3
  • Adi3
@Jhannybean
Adi3
  • Adi3
@AaronAndyson
Adi3
  • Adi3
@ganeshie8

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More answers

IrishBoy123
  • IrishBoy123
\(y = 2(n-2)^3 \) \((\dfrac{y}{2})^\frac{1}{3} = n-2\) etc
Adi3
  • Adi3
how did you do that
IrishBoy123
  • IrishBoy123
\(f(n) = 2(n-2)^3\) \(y = 2(n-2)^3\) yes?
IrishBoy123
  • IrishBoy123
\(\frac{y}{2} = (n-2)^3\) ok?
Adi3
  • Adi3
why would you do y/2, isn't it |dw:1443341707529:dw|
IrishBoy123
  • IrishBoy123
\[\dfrac{y}{2} = \dfrac{2(n-2)^3}{2}\]
IrishBoy123
  • IrishBoy123
you want the inverse, is that right, ie \(n = n(y)\)
IrishBoy123
  • IrishBoy123
\[\dfrac{y}{2} = \dfrac{2(n-2)^3}{2} \implies \dfrac{y}{2} = (n-2)^3\]
Adi3
  • Adi3
f(n) = 2(n-2)^3
Adi3
  • Adi3
we need to inverse this function
IrishBoy123
  • IrishBoy123
yes, indeed label the \(f(n)\) as y just because it makes it easier to write out. then we need to find and equation in form \(n = n(y) = f^{-1}(n)\) so \(y = 2(n-2)^3\) \(\frac{y}{2} = (n-2)^3\) like so.
Adi3
  • Adi3
but isn't it like n= 2(y-2)^3
Adi3
  • Adi3
i like this way
Adi3
  • Adi3
@Miracrown
Adi3
  • Adi3
@insa
IrishBoy123
  • IrishBoy123
ok let's go that way \(n= 2(y-2)^3\) \(\frac{n}{2} = (y-2)^3\)
Adi3
  • Adi3
alright, next
IrishBoy123
  • IrishBoy123
\[(\frac{n}{2})^{\frac{1}{3}} = y-2\]
IrishBoy123
  • IrishBoy123
yes?
Adi3
  • Adi3
wait wait, where did 1/3 come from
IrishBoy123
  • IrishBoy123
take cube root of each side
Adi3
  • Adi3
why cant it be |dw:1443342930062:dw|
IrishBoy123
  • IrishBoy123
\(a^3 = b\) \(a = b^{\frac{1}{3}} = \sqrt[3]{b}\) same thing
Adi3
  • Adi3
\[\frac{ \sqrt[3]{n} }{ 2}\]
Adi3
  • Adi3
lets go this way if you dont mind
IrishBoy123
  • IrishBoy123
if that way, than \[\large (\frac{n}{2})^{\frac{1}{3}} = \sqrt[3]{\frac{n}{2}}= y-2\]
Adi3
  • Adi3
yup, then \[\frac{ \sqrt[3]{n} }{ 2 } + 2\]
IrishBoy123
  • IrishBoy123
be careful the cube root must include the 1/2 its \[y=\sqrt[3]{\frac{ n} { 2 }} + 2\] NOT \[y=\frac{ \sqrt[3]{n} }{ 2 } + 2\]
Adi3
  • Adi3
yeah
Adi3
  • Adi3
so is that it
IrishBoy123
  • IrishBoy123
good!
Adi3
  • Adi3
one more question can i ask please
IrishBoy123
  • IrishBoy123
shoot but if it's a new question, new thread is best
Adi3
  • Adi3
find the inverse of g(x) = 1/x - 2
Adi3
  • Adi3
i know that x = 1/y - 2
Adi3
  • Adi3
@IrishBoy123
IrishBoy123
  • IrishBoy123
you like to do it this way \[x = \frac{1}{y} - 2 \] \[\frac{1}{y} = x + 2 \] \[y = \frac{1}{x+2}\]
Adi3
  • Adi3
and the inverse of -4x + 1 is \[\frac{ x-1 }{ 4 }\] right
IrishBoy123
  • IrishBoy123
|dw:1443344105564:dw|
Adi3
  • Adi3
alright thanks. no more questions

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