How do you find the inverse of # y = e^x-1/x#?

2 Answers
Oct 5, 2016

See below


To find the inverse of a given function, you need to isolate the #x# variable.

For example, if you have #y = 2x+7#, you can do the following:

#y = 2x+7 \to y-7 = 2x \to x=(y-7)/2#

In your case, it's not possible to isolate #x#, so you can't find an explicit expression for the inverse of your function.

The only thing you can do is something graphic, since the inverse of a function is obtained via symmetry with respect to the #y=x# line.

So, you can draw the graph of #e^x-1/x#, and then mirror it, and it would be the graph of the inverse function.

Oct 6, 2016

As of now, there is no way of finding the inverse.


Unfortunately, the distributive law is not applicable for inversion.

This means that if # y=f(x) + g(x), x is not the sum of the inverses of

f(x) and g(x).,

For example, if f(x) =x, g(x)=2 and y = f(x) + g(x) = x + 2,

inversely x= y-2..

The sum of the separate inverses is y+1/2.

Here, f(x) = #e^x and g(x)=-1/x#.

Despite that separate inverses are known as .ln y and -1/y,

we do not have a method to bring out the inverse of f+g.