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

Revision as of 11:51, 4 August 2020

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

Note that $f(x)=(x+2)^3-2$ . Thus $f(f(f(x)))=(x+2)^{27}-2$ . Setting this equal to zero and solving for $x$ , we have $\boxed{x=-2+\sqrt[27]{2}}$

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-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^2+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>.
+Note that <math>f(x)=(x+2)^3-2</math>. Thus <math>f(f(f(x)))=(x+2)^{27}-2</math>. Setting this equal to zero and solving for <math>x</math>, we have <math>\boxed{x=-2+\sqrt[27]{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 ==

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