Inverse function question (1 Viewer)

[rsingh]

Find the inverse function of y = x - e^-x.

I've tried doing it but can't seem to get "y" on its own once I interchange the variables. Any help would be appreciated!

Thanks.

 

[haboozin]

is this an actual question or did u just think of it to see if it is possible...?


if u draw y = x - e^-x and mirror it though y = x it is something like y = x - lnx

 

[Slidey]

It's nothing like y=x-lnx

y=x+e^x is a mirror image of it in one way. But not about the line y=x.

 

[lucifel]

when you start mixing exponents and algberaic expressions, it gets close to impossible to simply use algebraic manilpulations to change subject. y = x + e^x or any variant of that is one of those I think. Did you just make up the question? Or was it actually a question in a textbook? Because it does look like it is too hard to change the subject to express the inverse function as a function explictiy of x.

 

[Slidey]

Indeed, the transcendental nature of e^x should dictate it is impossible.

EDIT: To clarify: Impossible with the high school maths

 

[who_loves_maths]

... this looks like a function of a type similar to or associated with the product-log function, which is notorious (among other things) for having an inverse , which is a function, but that cannot be expressed in terms of elementary functions.

i suspect you simply made that question up yourself rsingh? (not that there's anything wrong with that )
or maybe you're trying to find the gradient of a particular point on the inverse function and thought that the only way to do that is by finding the actual inverse function (which you don't have to)?


Edit: finding the gradient of the tangent to a point on an inverse curve is a common mathematics question at the HSC (3u) level, which is why i mentioned it above.

 

[Slidey]

How would find the tangent if you didn't know the function?

I thought of a way but it's rather labourious and not exact.

E.g.: Find the tangent to the inverse of the curve y=x-1/e^x at the point x=a on the inverse curve.

dx/dy=1+1/e^y [EDIT: Derivative is dy/dx=1/(1+1/e^y) - I forgot I'd already swapped the variables.]
solve a=x-1/e^x for x, now use this x value for your y value on the inverse function's derivative so that you can find the gradient, then use point-gradient formula.

The non-exactness comes from the transcendental equation a=x-1/e^x.

 

[who_loves_maths]

Originally Posted by Slide_Rule
How would you find the tangent if you didn't know the function?

I thought of a way but it's rather labourious and not exact.

E.g.: Find the tangent to the inverse of the curve y=x-1/e^x at the point x=a on the inverse curve.

dx/dy=1+1/e^y
solve a=x-1/e^x for x, now use this x value for your y value on the inverse function's derivative so that you can find the gradient, then use point-gradient formula.

The non-exactness comes from the transcendental equation a=x-1/e^x.

hi Slide,

no you don't need the inverse function (in its explicit form) to find the gradient of the tangent to any point on the inverse curve.
however, what you do need is for the question to ask you to find the gradient of the tangent at a point where you are given the y-coordinate and not the 'x'.
[note: this requirement is only applicable to functions whose inverse you cannot express with 'y' as the subject, otherwise either the 'x' or 'y' coordinates will suffice.]

with your example:

let's say the y-coordinate is y=b ; ie. the point is P(a, b)

now, with the original function y = x + e^(-x) ; dy/dx = 1 - e^(-x)

ie. dx/dy = 1/(1 - e^(-x)) ---> dx/dy is the inverse derivative of dy/dx {for obvious reason which i'm sure you can see} ;

now, in the inverse function, the 'x' and 'y' changes place so the 'x's in the inverse derivative becomes the 'y's...

hence, the gradient of the tangent at any point on the inverse function is simply:
dy/dx = 1/(1 - e^(-y))

so in this case, the tangent is given by:
y - b = (x - a)/(1 - e^(-b)) ---> y = (x -a)/(1 - e^(-b)) + b

 

[Slidey]

So you think they would give you the y value instead?

Giving the you x value would be an interesting question. It would probably need to be structured such that you find an approximation to the root of the transcendental equation it produces, though, before hand (through Newton's or bisection method).

 

[who_loves_maths]

Originally Posted by Slide_Rule
So you think they would give you the y value instead?

yes i'm certain of it... they will not give you the 'x' value, only the 'y' one.

Originally Posted by Slide_Rule
Giving the you x value would be an interesting question. It would probably need to be structured such that you find an approximation to the root of the transcendental equation it produces, though, before hand (through Newton's or bisection method).

uhuh, if you're only given the 'x' coordinate, then by 3u techniques only an estimate for the gradient is obtainable.

 

[rsingh]

Hey guys, sorry for my late reply.

No it's a question I found in a past trial paper, they ask you to find the inverse function and sketch them both.
I can't remember which trial paper it was from, but that was definetly it.

So is there a logical way of doing this question, or is it meant to be an extremely difficult question?
Sorry I'm not following your methods involving tangents and the gradient.

 

[who_loves_maths]

Originally Posted by rsingh
No it's a question I found in a past trial paper, they ask you to find the inverse function and sketch them both.
I can't remember which trial paper it was from, but that was definetly it.

So is there a logical way of doing this question, or is it meant to be an extremely difficult question?

you don't need to find the inverse function in order to draw a sketch of its graph - you just reflect the original function in y=x.

however, if the question indeed asked you to find the inverse function, then i'd say that the question is vastly misguided - since the mathematics required to do this does not appear until second year maths at university...

so in reply to your "So is there a logical way of doing this question, or is it meant to be an extremely difficult question?" - Yes, there is (always) a logical way, and Yes it is extremely difficult at this stage.

 

[rsingh]

Well, thanks for your assistance guys. And you're right, the question only asks to sketch the graph. Sorry, I thought you needed to find the equation of the inverse to do it.

So with which inverse fn graphs can we just reflect it in y=x? All inverse functions, or just with exponential and logs?

 

[shafqat]

All. Tis the definition.

 

[rsingh]

Oh, thanks for that. I didn't know that was the definition.
But we just started the topic - its our last.

Thanks for your help!

 

[Slidey]

Just a note:

The inverse function can in fact be expressed as x=y-1/e^y. While you haven't isolated y=f(x), it would be better than nothing in an exam, as it DOES define the inverse function you are looking for.

 

[who_loves_maths]

^ good point. if a question simpy says "find the inverse function" then you'll get marks for just leaving it like that.