If g(x) = 6 + x + e^x, find g^-1of (7).

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If g(x) = 6 + x + e^x, find g^-1of (7).

Mathematics
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At vero eos et accusamus et iusto odio dignissimos ducimus qui blanditiis praesentium voluptatum deleniti atque corrupti quos dolores et quas molestias excepturi sint occaecati cupiditate non provident, similique sunt in culpa qui officia deserunt mollitia animi, id est laborum et dolorum fuga. Et harum quidem rerum facilis est et expedita distinctio. Nam libero tempore, cum soluta nobis est eligendi optio cumque nihil impedit quo minus id quod maxime placeat facere possimus, omnis voluptas assumenda est, omnis dolor repellendus. Itaque earum rerum hic tenetur a sapiente delectus, ut aut reiciendis voluptatibus maiores alias consequatur aut perferendis doloribus asperiores repellat.

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try x=0 see if \[g(0)=7\] then you will have your answer
you really have to guess this one. no computing it using algebra
ln(x-6)/2 gives me 0/2 or zero is this right

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Plug g(x)=7, and solve for x.
that is not the inverse. you would have to solve \[x=6+y+e^y\] for y. good luck!
yes but for this one you have to guess. you check that \[g(0)=6+0+e^0=6+1=7\] so \[g^{-1}(7)=0\]
don't try to find \[g^{-1}(x)\] using algebra
hello anwar. seen quite a spate of wrong answers here tonight
not from you of course
Yeah, Plugging \(7\) for g(x) gives: \(x+e^x=1\), then \(x=0\) is the only solution. Hence \(g^{-1}(7)=0\).
thanks
exactly but you solved the last one by inspection yes?
Yes! But, it's not hard to see it. It can be easily solved by graphing as well.
And hello satellite :)

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