Bloody Roots

Algebra Level 3

Consider a function $f(x)$ such that $f(2)=f(6)=0$ . Evaluate the sum of all right sentences.

(1) $f(x) = k(x^2 - 8x + 12), k \in \mathbb{Z}$

(2) $f(1) \cdot f(7) \geq 0$

(4) $f(f(2)) = f(f(6))$

(8) $f(4) \in \mathbb{C}$

(16) If $\displaystyle k^{ \displaystyle k^{ ( \cdots) }} = 2$ , then $f(4 \pm k^2) = 0$ .

(32) $f^{(2014)}(x) = k, k \in \mathbb{R}$

1 solution

Guilherme Dela Corte
May 17, 2014

(1) - Consider the function $f(x) = \log((x-2)(x-6) + 1)$ . Notice it yields our desired roots without being a quadratic.

(2) - Consider the function $f(x) = (x-2)^2 (x-6)$ . We have that $f(1) \cdot f(7) = -5$ has a negative value.

(4) - Consider the function $f(x) = \dfrac{(x-2)(x-6)}{x}$ . $f(0)$ is not defined, and the equality does not hold.

(8) - Consider the function $f(x) = \dfrac{(x-2)(x-6)}{x-4}$ . $f(4)$ is not defined, and thus it cannot be a complex number.

(16) - Solving the infinite exponentiation, we find $k = \sqrt{2}$ , and it yields $f(2) = f(6) = 0$ , which is true.

(32) - Consider the function $f(x) = \log((x-2)(x-6) + 1)$ . Its $n$ -th derivative, no matter how big $n$ is, never reaches a constant.

Wow....I limited my thinking to a quadratic :(

Tanya Gupta - 7 years ago

That's the catch I intended to make when designing this question!

Guilherme Dela Corte - 6 years, 11 months ago

For (1), (2), (4), (8), you might as just just define the function to be

$f(x) = (2,0), (6,0 ).$

Calvin Lin Staff - 6 years, 5 months ago

For 16, we can just substitute the given equation to itself to yield $k^2=2$ .

Kenny Lau - 6 years, 9 months ago