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how to find the inverse function of f(x)=κ-e^(2-x)+x  supposing that κ is a real number

Guest Jul 12, 2015

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 #2
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+8

Thanks a lot.I had solved the most difficult exercises and I was stuck here,because I did not realise that you could swap x and y without doing any calculations.Imagine that I even know what I have to do next in this exercise,but my only problem was finding f^-1(x).(I had already found that f was 1-1.I am inexcusable for getting stuck here,XDDDD.Thanks man

Guest Jul 12, 2015
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 #1
avatar+91233 
+5

f(x)=κ-e^(2-x)+x

 

Here is a graph of f(x)

https://www.desmos.com/calculator/rr5rg1wjd3

 

since the mapping of x to y and y to x are both 1 to 1   I can just let  f(x)=y and then swap x and y over.

 

function

$$y=k-e^{(2-x)}+x$$

 

inverse function

$$\\x=k-e^{(2-y)}+y\\$$

 

I should make y the subject but I cannot see how to do that.     

Melody  Jul 12, 2015
 #2
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+8
Best Answer

Thanks a lot.I had solved the most difficult exercises and I was stuck here,because I did not realise that you could swap x and y without doing any calculations.Imagine that I even know what I have to do next in this exercise,but my only problem was finding f^-1(x).(I had already found that f was 1-1.I am inexcusable for getting stuck here,XDDDD.Thanks man

Guest Jul 12, 2015
 #3
avatar+91233 
+3

It is much more elegant to make x the subject before you do the swap.

AND it is also much less likely that you will make mistakes 

BUT if you are sure it is continuous and that each y maps only to one x then I think this way is alright :)

 

I am glad I could help :)

Melody  Jul 12, 2015

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