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Could someone help me on this question please: If g(x) = 3+x+e^x, find g^-1(4) Answer: g^-1(4) = 0
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I replaced all x's with y's and came up with this, x-3=y+e^y and I'm not sure how to solve for y from here
*all x's with y's and all y's with x's
You need to find \[g ^{-1} (4)\] This means you need to find x for which g(x)=4 You agree with this?
Yeah, that makes sense so far
Though how would I do that? Is that just plug and check?
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so \[4=3+x+e^x\] Yeah here if we plug x=0 We'll get 4 so x=0 is the answer
I think we can only plug and check:(
Oh okay, thanks for your help, I appreciate it.
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