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

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Answer 1 · 2016-10-05T20:14:55

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.

Answer 2 · 2016-10-06T04:17:14

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.