Here's the question you clicked on:
iop360
isolate y for the following equation:
\[x= -14e ^{2-y} + y^3 + 7y + 6\]
i don't think there is an elementary method to do that.
hmm the original equation asks to find the inverse function of \[-14e ^{2-x} + x^3 + 7x + 6 \]
which means swapping y for x and solving for y
and then find the equation of the line perpendicular to this inverse function at x = 14
correct... but isolating the y after you reversed it is very difficult. i guess all you could do here is give the function implicitly.
can you give the original problem? looks like you need to take derivative implicitly here.
differentiate both sides of the equation, find y' maybe
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
normal line btw, means the line w/ a negative reciprocal
aka perpendicular. And the answer to this is y = -33x+464
do you know how the derivatives (slopes) of inverse functions are related?
maybe could you state it?
i get \[\frac{ dy }{ dx } = \frac{ 1 }{ 14e ^{2-y} + 3y^2 + 7 }\]
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
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.
how do we find\[f ^{-1}(14)\]
since f(2)=14, then f-1(14)=2
wasnt 2 just used as a hypothetical
yeah, thats a challenge \(14 = -14e ^{2-x} + x^3 + 7x + 6\) solve for x to get x=2 is not easy
i found it by trial and error method
the 1st thing i thought is to eliminate e, by putting x=2 and voila!
so now u have a point, the derivative,can u find the slope of line....and then the equation?
yes that makes it way easier
*point and derivative of inverse function....
yw... also remember you're not looking for the tangent but the normal line
i got this because i have seen a similar question here, http://www.physicsforums.com/showthread.php?t=397539