Adi3
  • Adi3
Will medal please help. consider g(n) = -12-2n/3. Find g^-1(n)
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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katieb
  • katieb
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Adi3
  • Adi3
@misty1212
Adi3
  • Adi3
@ganeshie8
Adi3
  • Adi3
@Jhannybean

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More answers

Jhannybean
  • Jhannybean
Is your function : \(g(n) = \dfrac{-12-2n}{3}\)?
Adi3
  • Adi3
yes
Adi3
  • Adi3
we need to find the inverse.
Jhannybean
  • Jhannybean
So first graph it to see whether it's a one-to-one function, and it is. \[y = \frac{-12-2n}{3}\]Switch y and n \[n=\frac{-12-2y}{3}\]Now lets solve for y.
Jhannybean
  • Jhannybean
I let my function \(g(n) = y\)
Jhannybean
  • Jhannybean
What function do you get when you resolve for y?
Adi3
  • Adi3
3n + 12 = -2y
Jhannybean
  • Jhannybean
Good, and when you isolate y?
Adi3
  • Adi3
that's where i reached till, i don't know what to do after that
Jhannybean
  • Jhannybean
To isolate the variable y, what do you have to do to each side of the equation?
Adi3
  • Adi3
divide by 2 i guess
Jhannybean
  • Jhannybean
-2, and yes, you were partially right.
Adi3
  • Adi3
so 3n + 12/-2 = y
Jhannybean
  • Jhannybean
(3n + 12)/ -2 = y * It helps when you put parenthesis () around the numerator and denominator
Adi3
  • Adi3
can I ask one more question please????
Jhannybean
  • Jhannybean
We can simplify that function a bit by dividing both terms in the numerator by -2. ((3n)/-2) +(12/-2) = y
Jhannybean
  • Jhannybean
Sure, I can try helping.
Adi3
  • Adi3
ok, find the inverse of the equation f(x) = \[-\frac{ 4 }{ 7 }x - \frac{ 16 }{ 7 }\]
Jhannybean
  • Jhannybean
\[y=\frac{-4x-16}{7}\]Same process, we switch x and y. \[x=\frac{-4y-16}{7}\] and now we resolve for y.
Adi3
  • Adi3
7x +16 = -4x
Adi3
  • Adi3
right?
Jhannybean
  • Jhannybean
-4y*
Adi3
  • Adi3
ohh yeah
Jhannybean
  • Jhannybean
7x + 16 = -4y and now we've got to isolate y, just like before.
Adi3
  • Adi3
\[\frac{ 7x+16 }{ -4 }\]
Jhannybean
  • Jhannybean
That's right, and after resolving for y, we relabel it as \(f^{-1}(x)\)
Adi3
  • Adi3
ok, can i ask further questions if i have doubt in them please?
Jhannybean
  • Jhannybean
Sure, but im logging off for a while, although good luck on learning this!`
Adi3
  • Adi3
when will you come back

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