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

Find the maximum value of \(f_3(x)\).

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 \(x=1/2\).

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].

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 \(1\).

  1. Write down \(f_1(x)\). Calculate \(f_1(f_1(x))\) and \(f_1(f_1(f_1(x)))\).
\[f_1(x)=4x+1\] So \[\begin{align*} f_1(f_1(x))&=4(4x+1)+1\\ &=16x+5 \end{align*}\] and \[\begin{align*} f_1(f_1(f_1(x)))&=f_1[f_1(f_1(x))]\\ &=4(16x+5)+1\\ &=64x+21. \end{align*}\]

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.]

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 \(f_1^k(x)=4^kx+\dfrac{4^k-1}{3}\).

  1. Write down \(f_2(x)\).


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

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?

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 \(2^k\).