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OpenStudy (anonymous):
If g(x) = 6 + x + e^x, find g^-1of (7).
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OpenStudy (anonymous):
try x=0 see if \[g(0)=7\] then you will have your answer
OpenStudy (anonymous):
you really have to guess this one. no computing it using algebra
OpenStudy (anonymous):
ln(x-6)/2 gives me 0/2 or zero is this right
OpenStudy (anonymous):
Plug g(x)=7, and solve for x.
OpenStudy (anonymous):
that is not the inverse. you would have to solve \[x=6+y+e^y\] for y. good luck!
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OpenStudy (anonymous):
yes but for this one you have to guess. you check that \[g(0)=6+0+e^0=6+1=7\]
so \[g^{-1}(7)=0\]
OpenStudy (anonymous):
don't try to find \[g^{-1}(x)\] using algebra
OpenStudy (anonymous):
hello anwar. seen quite a spate of wrong answers here tonight
OpenStudy (anonymous):
not from you of course
OpenStudy (anonymous):
Yeah, Plugging \(7\) for g(x) gives: \(x+e^x=1\), then \(x=0\) is the only solution. Hence \(g^{-1}(7)=0\).
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OpenStudy (anonymous):
thanks
OpenStudy (anonymous):
exactly but you solved the last one by inspection yes?
OpenStudy (anonymous):
Yes! But, it's not hard to see it. It can be easily solved by graphing as well.
OpenStudy (anonymous):
And hello satellite :)
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