Question

isolate y for the following equation:

Answer 1

\[x= -14e ^{2-y} + y^3 + 7y + 6\]

Answer 2

i don't think there is an elementary method to do that.

Answer 3

hmm the original equation asks to find the inverse function of
\[-14e ^{2-x} + x^3 + 7x + 6 \]

Answer 4

which means swapping y for x and solving for y

Answer 5

and then find the equation of the line perpendicular to this inverse function at x = 14

Answer 6

correct... but isolating the y after you reversed it is very difficult. i guess all you could do here is give the function implicitly.

Answer 7

can you give the original problem? looks like you need to take derivative implicitly here.

Answer 8

differentiate both sides of the equation, find y' maybe

Answer 9

ok

Answer 10

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

Answer 11

normal line btw, means the line w/ a negative reciprocal

Answer 12

aka perpendicular.
And the answer to this is y = -33x+464

Answer 13

do you know how the derivatives (slopes) of inverse functions are related?

Answer 14

maybe could you state it?

Answer 15

i get
\[\frac{ dy }{ dx } = \frac{ 1 }{ 14e ^{2-y} + 3y^2 + 7 }\]

Answer 16

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

Answer 17

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.

Answer 18

how do we find\[f ^{-1}(14)\]

Answer 19

since f(2)=14, then f-1(14)=2

Answer 20

oh right

Answer 21

wasnt 2 just used as a hypothetical

Answer 22

yeah, thats a challenge
\(14 = -14e ^{2-x} + x^3 + 7x + 6\)
solve for x
to get x=2 is not easy

Answer 23

i found it by trial and error method

Answer 24

oh ok

Answer 25

the 1st thing i thought is to eliminate e, by putting x=2 and voila!

Answer 26

oh i see

Answer 27

so now u have a point, the derivative,can u find the slope of line....and then the equation?

Answer 28

yes that makes it way easier

Answer 29

thanks to both of you

Answer 30

*point and derivative of inverse function....

Answer 31

welcome ^_^

Answer 32

yw... also remember you're not looking for the tangent but the normal line

Answer 33

i got this because i have seen a similar question here,
http://www.physicsforums.com/showthread.php?t=397539

Answer 34

i see