1. Problem b) Find the value of $p$ for which the function $$f(x) = \begin{cases} \frac{\sqrt{1+px} - \sqrt{1-px}}{x}, & -1 \leq x < 0 \\ \frac{2x+1}{x-2}, & 0 \leq x \leq 1 \end{cases}$$ is continuous at $x=0$. 2. For continuity at $x=0$, the left-hand limit $$\lim_{x \to 0^-} f(x)$$ must equal the right-hand limit $$\lim_{x \to 0^+} f(x)$$ and the function's value at $0$. 3. Evaluate right-hand limit using $$f(0) = \frac{2(0)+1}{0-2} = \frac{1}{-2} = -\frac{1}{2}$$. 4. Find left-hand limit: $$\lim_{x \to 0^-} \frac{\sqrt{1+px} - \sqrt{1-px}}{x}$$. 5. Multiply numerator and denominator by conjugate: $$\frac{\sqrt{1+px} - \sqrt{1-px}}{x} \times \frac{\sqrt{1+px} + \sqrt{1-px}}{\sqrt{1+px} + \sqrt{1-px}} = \frac{(1+px)-(1-px)}{x(\sqrt{1+px} + \sqrt{1-px})} = \frac{2px}{x(\sqrt{1+px} + \sqrt{1-px})}$$. 6. Simplify canceling $x$: $$= \frac{2p}{\sqrt{1+px} + \sqrt{1-px}}$$. 7. Taking the limit as $x \to 0^-$: $$\lim_{x \to 0^-} \frac{2p}{\sqrt{1+0} + \sqrt{1-0}} = \frac{2p}{1+1} = \frac{2p}{2} = p$$. 8. For continuity, equate left limit, right limit, and function value at $0$: $$p = -\frac{1}{2}$$. Final answer: $$\boxed{p = -\frac{1}{2}}$$.