+14913 If \(f(x)=x^3+3x^2+3x+1 \), find \(f(f^{-1}(2010)) \)
Hello Guest!
\(f(x)=x^3+3x^2+3x+1 \)
\(f(x)=y=x^3+3x^2+3x+1\\ f(x)=y=(x-1)^3\\ f^{-1}(x)=x=(y-1)^3\\ f^{-1}(x)=y=\sqrt[3]{x}+1\)
\(f^{-1}(2010)=\sqrt[3]{2010}+1=\pm 12.6202+1\\ f^{-1}(2010)\in \{13.6302,-11.6202\}\)
\(f(f^{-1}(2010))=f(13.6302, -11.6202) \)
\((x-1)^3=(13.6302-1)^3=2010\)
\((x-1)^3=(-11.6302-1)^3=-2010\)
\(f(f^{-1}(2010))\in \{-2010,2010\}\)
!
+118609 I think they cancel each other out and the answer is just 2010
Looking at it a little more formally.
I think it means
\(2010=(x+1)^3\\ \sqrt[3]{2010}-1=x\\ f(\sqrt[3]{2010}-1)=(\sqrt[3]{2010}-1+1)^3=2010\)
+14913 Hi Melody!
I think:
In the term \(f(f^{-1}(2010)\) is 2010 the argument of the \(f^{-1}\).
Then applies:
\(f^{-1}(x)=f^{-1}(2100)=\sqrt[3]{x}+1=\sqrt[3]{2010}+1\\ \color{blue}f^{-1}(2100)\in\{-11.6202,13.6202\}\)
These values of the function \(f^{-1}\)are two arguments for the function f.
Please confirm whether this is the right or wrong idea.
Grüße
