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Mathematics 18 Online
OpenStudy (anonymous):

HHHEEEEEELLLPPPP INVERSE FUNCTIONS

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

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

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

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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