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

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

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

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

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

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

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

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

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

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

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

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

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

hartnn Group TitleBest ResponseYou've already chosen the best response.0
i found it by trial and error method
 one year 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!
 one year 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?
 one year ago

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

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

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

hartnn Group TitleBest ResponseYou've already chosen the best response.0
welcome ^_^
 one year 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
 one year 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
 one year ago
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