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iop360 Group TitleBest ResponseYou've already chosen the best response.1
\[x= 14e ^{2y} + y^3 + 7y + 6\]
 2 years ago

ByteMe Group TitleBest ResponseYou've already chosen the best response.0
i don't think there is an elementary method to do that.
 2 years ago

iop360 Group TitleBest ResponseYou've already chosen the best response.1
hmm the original equation asks to find the inverse function of \[14e ^{2x} + x^3 + 7x + 6 \]
 2 years ago

iop360 Group TitleBest ResponseYou've already chosen the best response.1
which means swapping y for x and solving for y
 2 years ago

iop360 Group TitleBest ResponseYou've already chosen the best response.1
and then find the equation of the line perpendicular to this inverse function at x = 14
 2 years ago

ByteMe Group TitleBest ResponseYou've already chosen the best response.0
correct... but isolating the y after you reversed it is very difficult. i guess all you could do here is give the function implicitly.
 2 years ago

ByteMe Group TitleBest ResponseYou've already chosen the best response.0
can you give the original problem? looks like you need to take derivative implicitly here.
 2 years ago

iop360 Group TitleBest ResponseYou've already chosen the best response.1
differentiate both sides of the equation, find y' maybe
 2 years ago

iop360 Group TitleBest ResponseYou've already chosen the best response.1
let \[f(x) = 14e ^{2x} + x^3 + 7x + 6\] find an equation of the normal line to the graph of f1(x) at the point on the graph where x=14
 2 years ago

iop360 Group TitleBest ResponseYou've already chosen the best response.1
normal line btw, means the line w/ a negative reciprocal
 2 years ago

iop360 Group TitleBest ResponseYou've already chosen the best response.1
aka perpendicular. And the answer to this is y = 33x+464
 2 years ago

ByteMe Group TitleBest ResponseYou've already chosen the best response.0
do you know how the derivatives (slopes) of inverse functions are related?
 2 years ago

iop360 Group TitleBest ResponseYou've already chosen the best response.1
maybe could you state it?
 2 years ago

iop360 Group TitleBest ResponseYou've already chosen the best response.1
i get \[\frac{ dy }{ dx } = \frac{ 1 }{ 14e ^{2y} + 3y^2 + 7 }\]
 2 years ago

iop360 Group TitleBest ResponseYou've already chosen the best response.1
my prof gave a hint saying we had to find a value for x that would make the \[e ^{2x} \] part equal to 0. dont know if that helps tho
 2 years ago

hartnn Group TitleBest ResponseYou've already chosen the best response.0
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 = f1(y). The slope of that function at that point is what you are looking for.
 2 years ago

iop360 Group TitleBest ResponseYou've already chosen the best response.1
how do we find\[f ^{1}(14)\]
 2 years ago

ByteMe Group TitleBest ResponseYou've already chosen the best response.0
since f(2)=14, then f1(14)=2
 2 years ago

iop360 Group TitleBest ResponseYou've already chosen the best response.1
wasnt 2 just used as a hypothetical
 2 years ago

hartnn Group TitleBest ResponseYou've already chosen the best response.0
yeah, thats a challenge \(14 = 14e ^{2x} + x^3 + 7x + 6\) solve for x to get x=2 is not easy
 2 years ago

hartnn Group TitleBest ResponseYou've already chosen the best response.0
i found it by trial and error method
 2 years ago

hartnn Group TitleBest ResponseYou've already chosen the best response.0
the 1st thing i thought is to eliminate e, by putting x=2 and voila!
 2 years ago

hartnn Group TitleBest ResponseYou've already chosen the best response.0
so now u have a point, the derivative,can u find the slope of line....and then the equation?
 2 years ago

iop360 Group TitleBest ResponseYou've already chosen the best response.1
yes that makes it way easier
 2 years ago

iop360 Group TitleBest ResponseYou've already chosen the best response.1
thanks to both of you
 2 years ago

hartnn Group TitleBest ResponseYou've already chosen the best response.0
*point and derivative of inverse function....
 2 years ago

ByteMe Group TitleBest ResponseYou've already chosen the best response.0
yw... also remember you're not looking for the tangent but the normal line
 2 years ago

hartnn Group TitleBest ResponseYou've already chosen the best response.0
i got this because i have seen a similar question here, http://www.physicsforums.com/showthread.php?t=397539
 2 years ago
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