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help please?????????????????????????????????
 

 Nov 25, 2020
 #1
avatar+10793 
+1

If \(f(x)=x^3+3x^2+3x+1 \), find \(f(f^{-1}(2010)) \)

 

Hello Guest!

 

\(f(x)=x^3+3x^2+3x+1 \)

\(f(x)=y=x^3+3x^2+3x+1\\ f(x)=y=(x-1)^3\\ f^{-1}(x)=x=(y-1)^3\\ f^{-1}(x)=y=\sqrt[3]{x}+1\)

\(f^{-1}(2010)=\sqrt[3]{2010}+1=\pm 12.6202+1\\ f^{-1}(2010)\in \{13.6302,-11.6202\}\)

\(f(f^{-1}(2010))=f(13.6302, -11.6202) \)

 

\((x-1)^3=(13.6302-1)^3=2010\)

\((x-1)^3=(-11.6302-1)^3=-2010\)

 

\(f(f^{-1}(2010))\in \{-2010,2010\}\)

laugh  !

 Nov 26, 2020
edited by asinus  Nov 26, 2020
 #2
avatar+112000 
+3

I think they cancel each other out and the answer is just 2010

 

Looking at it a little more formally.

I think it means

\(2010=(x+1)^3\\ \sqrt[3]{2010}-1=x\\ f(\sqrt[3]{2010}-1)=(\sqrt[3]{2010}-1+1)^3=2010\)

 Nov 26, 2020
 #3
avatar+10793 
+1

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

asinus  Nov 26, 2020
 #4
avatar+112000 
0

I don't think so.. Let's ask Alan or Heureka .....

Melody  Nov 27, 2020
 #5
avatar+31506 
+2

I see it this way:

 

 Nov 27, 2020
 #6
avatar+112000 
0

Thanks Alan, that is a much better way to present it.  :)

Melody  Nov 27, 2020

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