Can we compose these polynomial functions?

Can we compose these polynomial functions?

Question

Let $f_n(x)=(2+(-2)^n)x^2+(n+3)x+n^2$ where $n$ is a positive integer and $x$ is any real number.

1. Write down $f_3(x)$. Find the maximum value of $f_3(x)$. For what values of $n$ does $f_n(x)$ have a maximum value (as $x$ varies)? [Note you are not being asked to calculate the value of this maximum].
2. Write down $f_1(x)$. Calculate $f_1(f_1(x))$ and $f_1(f_1(f_1(x)))$. Find an expression, simplified as much as possible, for $f_1(f_1(f_1(...f_1(x))))$ where $f_1$ is applied $k$ times. [Here $k$ is a positive integer.]
3. Write down $f_2(x)$. The function $f_2(f_2(f_2(...f_2(x)))),$ where $f_2$ is applied $k$ times, is a polynomial in $x$. What is the degree of this polynomial?

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The green curve here is $y=(2+(-2)^n)x^2+(n+3)x+n^2$, where $n$ is a positive integer.

What happens as we vary $n$? Make sure you can explain algebraically the behaviour you can see here.

Solution

First,

$$\begin{align*} f_3(x)&=(2+(-2)^3)x^2+(3+3)x+3^2\\ &=-6x^2+6x+9. \end{align*}$$

To find the maximum value, we can complete the square:

$$\begin{align*} f_3(x)&=-6x^2+6x+9\\ &=-6\left(x-\frac{1}{2}\right)^2+\frac{3}{2}+9\\ &=-6\left(x-\frac{1}{2}\right)^2+\frac{21}{2}. \end{align*}$$

The maximum value is therefore $21/2$, achieved at $x=1/2$.

The graph of $y = f_n(x)$ will be a parabola – it will have a maximum if it points upwards.

So for $f_n$ to have a maximum we require the coefficient of $x^2$ in $f_n$ to be negative.

Consider $2 + (-2)^n$ – this will be negative exactly when $n$ is an odd number greater than $1$.

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$$f_1(x)=4x+1$$

So

$$\begin{align*} f_1(f_1(x))&=4(4x+1)+1\\ &=16x+5 \end{align*}$$

and

$$\begin{align*} f_1(f_1(f_1(x)))&=f_1[f_1(f_1(x))]\\ &=4(16x+5)+1\\ &=64x+21. \end{align*}$$

We have $f_1(x)=4(4(...4(4x+1)+1...)+1)+1$ The coefficient of $x$ is $4^k$. The constant term is $1+4+ ... +4^{k-1}.$ This is a geometric progression with common ratio $4$, so its sum is $1 \times \frac{4^k-1}{4-1}=\frac{4^k-1}{3}$ So $f_1^k(x)=4^kx+\dfrac{4^k-1}{3}$.

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$f_2(x)=6x^2+5x+4$.

If a question says ‘Write down…’ then that is just what we do — no justification is required.

Every time we apply $f_2$ to a polynomial, the degree of the polynomial is doubled. So $f_2^k$ is a polynomial of degree $2^k$.