how to factor a polynomial to the yth degree

How would I go about finding the inverse of the function f(x)=3+x+e^x

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I can rewrite the forumla interchanging the x and y variables to obtain

x = 3+y+e^y

but I'm not sure where to go from here.

thanks!

1st you should find x in terms of y as x=g(y), then switch x & y positions and check domain and range values.

Quote:

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Originally Posted by kingsolomonsgrave

How would I go about finding the inverse of the function f(x)=3+x+e^x

?


I can rewrite the forumla interchanging the x and y variables to obtain

x = 3+y+e^y

but I'm not sure where to go from here.

thanks!

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You will not be able to get an explicit equation y in terms of x, because of the fact that y is both "inside" a polynomial function and a transcendental function.

kingsolomonsgrave, where did you get this question from?

Quote:

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Originally Posted by Prove It

You will not be able to get an explicit equation y in terms of x, because of the fact that y is both "inside" a polynomial function and a transcendental function.

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thanks!

Quote:

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Originally Posted by Vlasev

kingsolomonsgrave, where did you get this question from?

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I got this from section 1.6 of James Stewart's 7th edition of Calculus

this one: http://www.amazon.com/Calculus-Trans.../dp/0538497904


the full question says

f(x)=3+x+e^x and then asks what f^-1(4) is

NOTE: by f^-1(4) I mean 'the inverse function of f at 4'

Quote:

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Originally Posted by kingsolomonsgrave

the full question says
f(x)=3+x+e^x and then asks what f^-1(4) is

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Looking at the derivative, it easy to see that function is one-to-one.
So thus .

It appears that your own solution is the best mirror image on y=x line:

http://mathhelpforum.com/attachment....1&d=1346260719

Quote:

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Originally Posted by Plato

Looking at the derivative, it easy to see that function is one-to-one.
So thus .

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Oh i see what the question was getting at now!