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Mathematics
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OpenStudy (anonymous):
HHHEEEEEELLLPPPP INVERSE FUNCTIONS
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OpenStudy (anonymous):
\[f(x)= -2(x+1)^2-4\]
TheSmartOne (thesmartone):
To find the inverse function we first switch x and y and then solve for y.
OpenStudy (jdoe0001):
same as you'd a linear one
do the variable "switcharoo" and solve for "y"
TheSmartOne (thesmartone):
^^ ditto
OpenStudy (anonymous):
naaaaaaaa id rather not
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OpenStudy (anonymous):
\[x= -2(y+1)^2-4\]
OpenStudy (jdoe0001):
yes
OpenStudy (anonymous):
i dont know what to do after that. Do i solve for y?
TheSmartOne (thesmartone):
yes.
OpenStudy (jdoe0001):
well... do you know how to simplify linear equations? is the same thing
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OpenStudy (jdoe0001):
say how would you simplify or solve for "a"
b = -2a -4 ?
OpenStudy (anonymous):
wouldnt you add 4\[x+4 = -2(x+1)^2\]
OpenStudy (jdoe0001):
well... yes... "y" on the right-side of course... so
\(\bf f(x)=y= -2(x+1)^2-4\qquad inverse\implies x=-2(y+1)^2-4\impliedby f^{-1}
\\ \quad \\
x+4=-2(y+1)^2\implies \cfrac{x+4}{-2}=(y+1)^2\quad \textit{ taking }\sqrt{\qquad }
\\ \quad \\
\sqrt{\cfrac{x+4}{-2}}=\sqrt{(y+1)^2}\implies \sqrt{\cfrac{x+4}{-2}}=(y+1)\implies ?\)
OpenStudy (anonymous):
\[\sqrt{(\frac{ x+4 }{ -2 })} -1\]
would it be that?
OpenStudy (anonymous):
@jdoe0001
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OpenStudy (jdoe0001):
yes
OpenStudy (anonymous):
is that the answer?
OpenStudy (jdoe0001):
yeap that's the "inverse relation"
mind you that is NOT a function.... but an inverse relationship expression
OpenStudy (anonymous):
alright, thank you. Could you help me with one more please?
OpenStudy (jdoe0001):
sure just post anew, thus if I dunno, someone else may know, and we revise each other
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