2014 UNM-PNM Statewide High School Mathematics Contest II Problems/Problem 2

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

2014 UNM-PNM Contest II (Problems • Answer Key • Resources)
Preceded by Problem 1	Followed by Problem 3
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2014 UNM-PNM Statewide High School Mathematics Contest II Problems/Problem 2

Contents

Problem

Solution

Solution 2

See also