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

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Nov 25, 2020

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

Hello Guest!

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

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

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

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

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

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

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

!

Nov 26, 2020
edited by asinus  Nov 26, 2020
#2
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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
+10793
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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
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I don't think so.. Let's ask Alan or Heureka .....

Melody  Nov 27, 2020
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+31506
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I see it this way:

Nov 27, 2020
#6
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Thanks Alan, that is a much better way to present it.  :)

Melody  Nov 27, 2020