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Additional Math Help HERE ONLY...!

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J.Darren:
https://studentforums.biz/index.php/topic,54.msg234150.html#msg234150

Dibss:

--- Quote from: J.Darren on June 07, 2010, 04:19:08 pm ---https://studentforums.biz/index.php/topic,54.msg234150.html#msg234150

--- End quote ---
Thank you so much! =]

BlackBunny103:

--- Quote from: J.Darren on June 07, 2010, 04:13:48 pm ---i ) Subsitute x = 0, the range would be greater than e^-1.

ii ) f(x) = e^(x+1)

y = e^(x+1)

ln y = (x+1) ln e

ln y = x + 1

ln x = y + 1

ln x - 1 = y

iii ) the domain of f-1(x) = the range of f(x), i.e. x > e^-1

--- End quote ---

Thanxx Darren ;). +rep
But I don't understand why from going   ln y = (x+1) ln e  to ln y = x + 1 why did you cancel lne but still keep ln e?
And why is the domain of f-1(x) = the range of f(x) ?
Thanxx again :D

J.Darren:

--- Quote from: BlackBunny103 on June 07, 2010, 04:24:12 pm ---Thanxx Darren ;). +rep
But I don't understand why from going   ln y = (x+1) ln e  to ln y = x + 1 why did you cancel lne but still keep ln e?
And why is the domain of f-1(x) = the range of f(x) ?
Thanxx again :D

--- End quote ---
I pretained ln e there just so that you can comprehend what exactly has happened :D

Bear in mind that the inverse of every function is the reflection of the original function on the line y = x, as we know plugging any value of the range of the original function in a one-to-one equation into its inverse would get us back to where we strated from ...

BlackBunny103:
Guys would you please help me out with these permutation & combination questions?
Thanxx ;)

My exam is in 4 hours :-\ ...

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