So Many Functions!

Algebra Level 3

Let there be functions a ( x ) , b ( x ) , a(x), b(x), and c ( x ) c(x) :

a ( x ) = x ² + 4 x + 4 a(x) = x² +4x +4 , b ( x ) = 1 x + 4 x ² + 4 b(x)= \frac{1}{x⁴ +4x² +4} , c ( x ) = x ² c(x)=x²

Then, ( a ¹ b c ¹ ) ( x ) (a⁻¹○b○c⁻¹)(x) can be expressed as two different functions, S ( x ) S₁(x) or S ( x ) S₂(x) .

Then, let d ( x ) = S ( x ) ² + S ( x ) ² d(x)= S₁(x)² + S₂(x)² . Find the sum of all complex zeroes of d ( x ) d(x) .

Definitions:

( f g h ) ( x ) (f○g○h)(x) means f ( g ( h ( x ) ) ) f(g(h(x)))

f ¹ ( x ) f⁻¹(x) means the inverse of the function f ( x ) f(x) .


The answer is -4.

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1 solution

Tom Engelsman
Aug 9, 2020

Let a 1 ( x ) = x 2 a^{-1}(x) = \sqrt{x} -2 and c 1 ( x ) = x c^{-1}(x) = \sqrt{x} . The three-fold composition computes to:

a 1 ( b ( c 1 ( x ) ) ) = 1 x + 2 2 S 1 ( x ) = 1 x + 2 , S 2 ( x ) = 2. a^{-1}(b(c^{-1}(x))) = \frac{1}{x+2} - 2 \Rightarrow S_{1}(x) = \frac{1}{x+2}, S_{2}(x) = -2.

Taking d ( x ) = S 1 2 ( x ) + S 2 2 ( x ) = 1 ( x + 2 ) 2 + 4 = 0 d(x) = S_{1}^{2}(x) + S_{2}^{2}(x) = \frac{1}{(x+2)^{2}} + 4 = 0 , the complex zeros compute to:

1 x + 2 = ± 2 i x = 2 ± 1 2 i \frac{1}{x+2} = \pm 2i \Rightarrow x = -2 \pm \frac{1}{2} i

and sum to 4 . \boxed{-4}.

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