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Mathematics 18 Online
OpenStudy (anonymous):

Find an explicit formula for f^(-1) and use it to graph f^(-1), f, and the line y=x on the same screen. To check your work, see whether the graphs of f and f^(-1) are reflections about the line. f(x)=2-e^(x)

jimthompson5910 (jim_thompson5910):

f(x)=2-e^(x) y=2-e^(x) Swap x and y to get x = 2 - e^y Now solve for y x-2 = -e^y e^y = -x + 2 I'll let you finish

OpenStudy (anonymous):

can you explain what I'm supposed to do.. I don't even understand what its asking.

jimthompson5910 (jim_thompson5910):

Do you know how to solve e^y = -x+2 for y?

OpenStudy (anonymous):

no lol.

jimthompson5910 (jim_thompson5910):

when we want to isolate exponents, what do we use?

OpenStudy (anonymous):

no idea....

OpenStudy (anonymous):

chain rule..?

jimthompson5910 (jim_thompson5910):

to undo addition, we use subtraction (or vice versa) to undo division, we use multiplication (or vice versa) to undo exponentiation, we use logs (or vice versa) etc

jimthompson5910 (jim_thompson5910):

so to solve 2^x = 16, we use logs

jimthompson5910 (jim_thompson5910):

what do you get when you solve e^y = -x+2 for y? (use logs)

OpenStudy (anonymous):

do i use ln?

jimthompson5910 (jim_thompson5910):

yes, ln (LN) is the natural log (it's a log with base 'e')

OpenStudy (anonymous):

ln(-x+2)?

jimthompson5910 (jim_thompson5910):

so y = ln(-x + 2)

jimthompson5910 (jim_thompson5910):

the last thing to do is replace y with \(\Large f^{-1}(x)\) to get the inverse \[\Large f^{-1}(x) = \ln(-x+2)\]

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