Art of Problem Solving
AoPS Online
Math texts, online classes, and more
for students in grades 5-12.
Visit AoPS Online ‚
Books for Grades 5-12 Online Courses
Beast Academy
Engaging math books and online learning
for students ages 6-13.
Visit Beast Academy ‚
Books for Ages 6-13 Beast Academy Online
AoPS Academy
Small live classes for advanced math
and language arts learners in grades 2-12.
Visit AoPS Academy ‚
Find a Physical Campus Visit the Virtual Campus

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

(→‎Solution)
(→‎Solution)
Line 3: Line 3:
  
 
If <math>f(x) = x^3 + 6x^2 + 12x + 6</math>, solve the equation <math>f(f(f(x))) = 0.</math>
 
If <math>f(x) = x^3 + 6x^2 + 12x + 6</math>, solve the equation <math>f(f(f(x))) = 0.</math>
−
 
−
== 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 x = \sqrt[3]{2+\sqrt[9]{2}}</math>.
 
  
 
==Solution 2==
 
==Solution 2==

Revision as of 01:15, 4 August 2020

Problem

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

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 (Problems • Answer Key • Resources)
Preceded by
Problem 1 		Followed by
Problem 3
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10
All UNM-PNM Problems and Solutions
Retrieved from "https://artofproblemsolving.com/wiki/index.php?title=2014_UNM-PNM_Statewide_High_School_Mathematics_Contest_II_Problems/Problem_2&oldid=130529"