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anonymous
 5 years ago
If g(x) = 6 + x + e^x, find g^1of (7).
anonymous
 5 years ago
If g(x) = 6 + x + e^x, find g^1of (7).

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anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0try x=0 see if \[g(0)=7\] then you will have your answer

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0you really have to guess this one. no computing it using algebra

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0ln(x6)/2 gives me 0/2 or zero is this right

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Plug g(x)=7, and solve for x.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0that is not the inverse. you would have to solve \[x=6+y+e^y\] for y. good luck!

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0yes but for this one you have to guess. you check that \[g(0)=6+0+e^0=6+1=7\] so \[g^{1}(7)=0\]

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0don't try to find \[g^{1}(x)\] using algebra

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0hello anwar. seen quite a spate of wrong answers here tonight

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0not from you of course

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Yeah, Plugging \(7\) for g(x) gives: \(x+e^x=1\), then \(x=0\) is the only solution. Hence \(g^{1}(7)=0\).

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0exactly but you solved the last one by inspection yes?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Yes! But, it's not hard to see it. It can be easily solved by graphing as well.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0And hello satellite :)
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