Problem Analysis #28

To find the number of intersection points of the curves $y = f(x)$ and $y = g(x)$ for $x \ge 0$, we first simplify the function $f(x)$ using $a = 3/5$:

The function $g(x)$ is a periodic triangular-wave function with period $2$ and range $[0, 1]$. Specifically, for $x \ge 0$:

• For $x \in [0, 1]$, $g(x) = x$ (increasing from $0$ to $1$)
• For $x \in [1, 2]$, $g(x) = 2-x$ (decreasing from $1$ to $0$)
• For $x \in [2, 3]$, $g(x) = x-2$ (increasing from $0$ to $1$)
• For $x \in [3, 4]$, $g(x) = 4-x$ (decreasing from $1$ to $0$)
• For $x \ge 4$, $g(x)$ continues this periodic pattern.

Since $0 \le g(x) \le 1$ for all $x$, any intersection point must satisfy $f(x) \le 1$. We solve for the range of $x$ where $f(x) \le 1$:

For $x \ge 0$, this gives 0 \le x \le 1 + $$\frac{\sqrt{80}}{3}$$ $$\approx$$ 3.98. Thus, there are no intersection points for $x \ge 4$ because $f(x) > 1$ in this region.

We now analyze the intersections in each unit interval for $0 \le x \le 4$:

1. Interval $[0, 1]$:

   • $f(x)$ is decreasing with $f(0) = \frac{49}{120} \approx 0.41$ and $f(1) = \frac{1}{3} \approx 0.33$.
   • $g(x) = x$ is increasing with $g(0) = 0$ and $g(1) = 1$.
   • Since $f(0) > g(0)$ and $f(1) < g(1)$, there is exactly $1$ intersection point.

2. Interval $[1, 2]$:

   • $f(x)$ is increasing with $f(1) = \frac{1}{3} \approx 0.33$ and $f(2) = \frac{49}{120} \approx 0.41$.
   • $g(x) = 2-x$ is decreasing with $g(1) = 1$ and $g(2) = 0$.
   • Since $f(1) < g(1)$ and $f(2) > g(2)$, there is exactly $1$ intersection point.

3. Interval $[2, 3]$:

   • $f(x)$ is increasing with $f(2) = \frac{49}{120} \approx 0.41$ and $f(3) = \frac{19}{30} \approx 0.63$.
   • $g(x) = x-2$ is increasing with $g(2) = 0$ and $g(3) = 1$.
   • Solving $f(x) = g(x)$ on this interval yields the quadratic equation: $$\frac{3}{40}(x-1)^2 + \frac{1}{3} = x-2 \implies 9(x-1)^2 - 120(x-1) + 160 = 0.$$ The roots for $x$ are $x \approx 2.50$ and $x \approx 12.83$. Only $x \approx 2.50$ lies in $[2, 3]$, giving exactly $1$ intersection point.

4. Interval $[3, 4]$:

   • $f(x)$ is increasing with $f(3) = \frac{19}{30} \approx 0.63$ and $f(4) = \frac{121}{120} \approx 1.01$.
   • $g(x) = 4-x$ is decreasing with $g(3) = 1$ and $g(4) = 0$.
   • Since $f(3) < g(3)$ and $f(4) > g(4)$, there is exactly $1$ intersection point.

Summing the intersections across all intervals, we find a total of $1 + 1 + 1 + 1 = 4$ intersection points.

\boxed{4}