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)
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
can you explain what I'm supposed to do.. I don't even understand what its asking.
Do you know how to solve e^y = -x+2 for y?
no lol.
when we want to isolate exponents, what do we use?
no idea....
chain rule..?
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
so to solve 2^x = 16, we use logs
what do you get when you solve e^y = -x+2 for y? (use logs)
do i use ln?
yes, ln (LN) is the natural log (it's a log with base 'e')
ln(-x+2)?
so y = ln(-x + 2)
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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