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iop360
 3 years ago
isolate y for the following equation:
iop360
 3 years ago
isolate y for the following equation:

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

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

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

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

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

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

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

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

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

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

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

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

iop360
 3 years ago
Best ResponseYou've already chosen the best response.1maybe could you state it?

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

iop360
 3 years ago
Best ResponseYou've already chosen the best response.1my 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

hartnn
 3 years ago
Best ResponseYou've already chosen the best response.0no, 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.

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

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

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

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

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

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

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

iop360
 3 years ago
Best ResponseYou've already chosen the best response.1yes that makes it way easier

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

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

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