Given the function f(x)=(x-9)^2 where x is less than or equal to 9 , find f^-1(x)
lest do it correctly
\[x=(y-9)^2\] \[y-9=\pm\sqrt{x}\] \[y=\pm\sqrt{x}+9\]
and now we have to decide plus or minus for the root, because otherwise we do not have a function
okay i got it! thank you so much :) I have one more problem and I have been working on this for three and a half hours basically. It's my recent one on the newsfeeed. Could you pleaseee help me.
since we choose the domain of f to be \[x\leq9\] we have to choose range of \[f^{-1}\] as \[f^{-1}(x)\leq 9\] \[\pm\sqrt{x}+9\leq 9\] \[\pm\sqrt{x}\leq 0\] which means we are picking \[-\sqrt{x}\]
okay i got it! thank you so much :) I have one more problem and I have been working on this for three and a half hours basically. It's my recent one on the newsfeeed. Could you pleaseee help me.
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sure
okay i got it! thank you so much :) I have one more problem and I have been working on this for three and a half hours basically. It's my recent one on the newsfeeed. Could you pleaseee help me.
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