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

(Created page with "== Problem == If <math>f(x) = x^3 + 6x^2 + 12x + 6</math>, solve the equation <math>f(f(f(x))) = 0.</math> == Solution == == See also == {{UNM-PNM Math Contest box|year=2014|...")
 
(Solution)
Line 5: Line 5:
  
 
== Solution ==
 
== Solution ==
 +
We can write <math>f(x) = x^3 + 6x^2 + 12x + 6 = (x+2)^3 -2</math>. This means that <math>f(x) = 0 \Rightarrow x = \sqrt[3]{2} - 2</math>.
 +
 +
Working backwards, <math>f(f(f(x))) = 0 \Rightarrow f(f(x)) = \sqrt[3]{2} - 2</math>.
 +
 +
Then, <math>f(f(x)) = (f(x)-2)^3 - 2 = \sqrt[3]{2} - 2 \Rightarrow f(x) = \sqrt[9]{2}</math>.
 +
 +
Finally, <math>f(x) = (x-2)^3 - 2 = \sqrt[9]{2} \Rightarrow \sqrt[3]{2+\sqrt[9]{2}}</math>.
  
 
== See also ==
 
== See also ==

Revision as of 19:32, 25 July 2016

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

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
Invalid username
Login to AoPS