Difference between revisions of "2014 UNM-PNM Statewide High School Mathematics Contest II Problems/Problem 2"

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Finally, <math>f(x) = (x-2)^3 - 2 = \sqrt[9]{2} \Rightarrow x = \sqrt[3]{2+\sqrt[9]{2}}</math>.
 
Finally, <math>f(x) = (x-2)^3 - 2 = \sqrt[9]{2} \Rightarrow x = \sqrt[3]{2+\sqrt[9]{2}}</math>.
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==Solution 2==
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Note that <math>f(x)=(x+2)^3-2</math>. Thus, <math>f^{n}(x)=(x+2)^{3n}-2</math>. In the question, <math>n=3</math>, so we have <math>(x+2)^{27}-2=0</math>. Solving, <math>\boxed{x=-2+\sqrt[27]{2}}</math>
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~yofro
  
 
== See also ==
 
== See also ==

Revision as of 23:12, 3 August 2020

Problem

If $f(x) = x^3 + 6x^2 + 12x + 6$, solve the equation $f(f(f(x))) = 0.$

Solution

We can write $f(x) = x^3 + 6x^2 + 12x + 6 = (x+2)^3 -2$. This means that $f(x) = 0 \Rightarrow x = \sqrt[3]{2} - 2$.

Working backwards, $f(f(f(x))) = 0 \Rightarrow f(f(x)) = \sqrt[3]{2} - 2$.

Then, $f(f(x)) = (f(x)-2)^3 - 2 = \sqrt[3]{2} - 2 \Rightarrow f(x) = \sqrt[9]{2}$.

Finally, $f(x) = (x-2)^3 - 2 = \sqrt[9]{2} \Rightarrow x = \sqrt[3]{2+\sqrt[9]{2}}$.

Solution 2

Note that $f(x)=(x+2)^3-2$. Thus, $f^{n}(x)=(x+2)^{3n}-2$. In the question, $n=3$, so we have $(x+2)^{27}-2=0$. Solving, $\boxed{x=-2+\sqrt[27]{2}}$

~yofro

See also

2014 UNM-PNM Contest II (ProblemsAnswer KeyResources)
Preceded by
Problem 1
Followed by
Problem 3
1 2 3 4 5 6 7 8 9 10
All UNM-PNM Problems and Solutions
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