Given that \(f(x) = x^2\) for \(-a < x \leq a\), and that \(f(x)\) is periodic with period \(2a\),

  1. sketch the graph of \(f(x)\) for \(-3a\leq x\leq 3a\),

Let’s start by sketching the graph of \(f(x)=x^2\) for values of \(x\) between \(-a\) and \(a\).

The graph. It is an vertex-down parabola.

Since the function is periodic with period \(2a\), the graph of the function for \(-3a\leq x\leq 3a\) will consist of three copies of the one above, together with a single point (for \(x = -3a\)).

One copy will be for \(-3a < x\leq -a\), one for \(-a < x \leq a\), and the last one for \(a < x\leq 3a\). Therefore the graph of \(f(x)\) for \(-3a\leq x\leq 3a\) is

The graph of the periodic function in the range described. In each of the three regions it resembles the first graph.

We might notice that the curve is continuous - there’s no jump from one repeat of the curve to the next, which means we can draw it without taking our pencil off the paper.

  1. find the general solution of the equation \(f(x)=\frac{1}{4} a^2\).

Let’s add to our graph the line \(y=\frac{1}{4}a^2\) to help us see the solution.

The graph as before but with the line y = a squared over 4 drawn on. It intersects the curve twice in each parabola region.

We need to solve the equation \(f(x)=\frac{1}{4} a^2\) when \(-a<x\leq a\). We have \[\frac{1}{4} a^2 = x^2 \iff x^2=\left(\frac{1}{2}a\right)^2 \iff x=\pm \frac{1}{2}a.\]

Since the function is periodic with period \(2a\), we have \(f(x)= f(x+2ak)\) for any integer \(k\) and any \(-a < x\leq a\).

So since the only solutions of \(f(x)=\dfrac{1}{4} a^2\) for \(-a<x\leq a\) are \(x=\pm \dfrac{1}{2}a\), then the solutions of \(f(x)=\dfrac{1}{4} a^2\) in general are \(\pm\dfrac{a}{2}+2ak\), where \(k\) takes integer values.