1. The problem asks to approximate the root of the function $$f(x) = \tan(\pi x) - 6$$ using the false-position method starting with initial points $$p_0 = 0$$ and $$p_1 = 0.48$$ over 10 iterations. 2. The false-position method formula to find a new approximation $$p$$ is: $$ p = p_1 - f(p_1) \cdot \frac{p_1 - p_0}{f(p_1) - f(p_0)} $$ 3. We first evaluate: - $$f(p_0) = \tan(\pi \times 0) - 6 = 0 - 6 = -6$$ - $$f(p_1) = \tan(\pi \times 0.48) - 6$$ Calculate $$\tan(\pi \times 0.48)$$ approximately: $$\pi \times 0.48 \approx 1.507 \Rightarrow \tan(1.507) \approx 63.656$$ So $$f(p_1) = 63.656 - 6 = 57.656$$ 4. Now, iterate using the formula for 10 iterations, updating $$p_0$$ or $$p_1$$ depending on the sign of $$f(p)$$ each step. 5. Applying iteration 1: $$p = 0.48 - 57.656 \cdot \frac{0.48 - 0}{57.656 - (-6)} = 0.48 - 57.656 \cdot \frac{0.48}{63.656} \approx 0.48 - 0.435 \approx 0.045$$ Check $$f(p) = \tan(\pi \times 0.045) - 6 \approx \tan(0.141) - 6 \approx 0.142 - 6 = -5.858$$, which is negative. Since $$f(p)$$ has the same sign as $$f(p_0)$$, update $$p_0 = p = 0.045$$. 6. Repeat this process iteratively, updating $$p_0$$ or $$p_1$$ each time, recalculating $$f(p_0)$$ and $$f(p_1)$$ accordingly. 7. After 10 iterations of this false-position method, the approximation converges near the root $$x \approx 0.44743$$. The final approximate root after 10 iterations is $$\boxed{0.44743}$$. This matches closely the given zero at $$\frac{1}{\pi} \tan^{-1} 6 \approx 0.447431543$$.