+0

# Inverse function

0
450
3

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

#2
+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
#1
+93683
+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
+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
#3
+93683
+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