anonymous
  • anonymous
isolate y for the following equation:
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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jamiebookeater
  • jamiebookeater
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anonymous
  • anonymous
\[x= -14e ^{2-y} + y^3 + 7y + 6\]
anonymous
  • anonymous
i don't think there is an elementary method to do that.
anonymous
  • anonymous
hmm the original equation asks to find the inverse function of \[-14e ^{2-x} + x^3 + 7x + 6 \]

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anonymous
  • anonymous
which means swapping y for x and solving for y
anonymous
  • anonymous
and then find the equation of the line perpendicular to this inverse function at x = 14
anonymous
  • anonymous
correct... but isolating the y after you reversed it is very difficult. i guess all you could do here is give the function implicitly.
anonymous
  • anonymous
can you give the original problem? looks like you need to take derivative implicitly here.
anonymous
  • anonymous
differentiate both sides of the equation, find y' maybe
anonymous
  • anonymous
ok
anonymous
  • anonymous
let \[f(x) = -14e ^{2-x} + x^3 + 7x + 6\] find an equation of the normal line to the graph of f-1(x) at the point on the graph where x=14
anonymous
  • anonymous
normal line btw, means the line w/ a negative reciprocal
anonymous
  • anonymous
aka perpendicular. And the answer to this is y = -33x+464
anonymous
  • anonymous
do you know how the derivatives (slopes) of inverse functions are related?
anonymous
  • anonymous
maybe could you state it?
anonymous
  • anonymous
i get \[\frac{ dy }{ dx } = \frac{ 1 }{ 14e ^{2-y} + 3y^2 + 7 }\]
anonymous
  • anonymous
my prof gave a hint saying we had to find a value for x that would make the \[e ^{2-x} \] part equal to 0. dont know if that helps tho
hartnn
  • hartnn
no, need to find the inverse function, u need to find y when x=14 lets say y=2 so u get \(f^{-1}(14)=2\) then (14,2) is on the graph of x = f-1(y). The slope of that function at that point is what you are looking for.
anonymous
  • anonymous
how do we find\[f ^{-1}(14)\]
anonymous
  • anonymous
since f(2)=14, then f-1(14)=2
anonymous
  • anonymous
oh right
anonymous
  • anonymous
wasnt 2 just used as a hypothetical
hartnn
  • hartnn
yeah, thats a challenge \(14 = -14e ^{2-x} + x^3 + 7x + 6\) solve for x to get x=2 is not easy
hartnn
  • hartnn
i found it by trial and error method
anonymous
  • anonymous
oh ok
hartnn
  • hartnn
the 1st thing i thought is to eliminate e, by putting x=2 and voila!
anonymous
  • anonymous
oh i see
hartnn
  • hartnn
so now u have a point, the derivative,can u find the slope of line....and then the equation?
anonymous
  • anonymous
yes that makes it way easier
anonymous
  • anonymous
thanks to both of you
hartnn
  • hartnn
*point and derivative of inverse function....
hartnn
  • hartnn
welcome ^_^
anonymous
  • anonymous
yw... also remember you're not looking for the tangent but the normal line
hartnn
  • hartnn
i got this because i have seen a similar question here, http://www.physicsforums.com/showthread.php?t=397539
anonymous
  • anonymous
i see

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