1. **State the problem:** Find the right-hand derivative of the piecewise function $$f(x) = \begin{cases} x^2 - 1, & x < 0 \\ 2x - 1, & x \geq 0 \end{cases}$$ at $$x=0$$. 2. **Recall the definition of the right-hand derivative:** The right-hand derivative at $$x=0$$ is given by $$f'_+(0) = \lim_{h \to 0^+} \frac{f(0+h) - f(0)}{h}$$ 3. **Evaluate $$f(0)$$:** Since $$0 \geq 0$$, use the second piece: $$f(0) = 2(0) - 1 = -1$$ 4. **Evaluate $$f(0+h)$$ for $$h > 0$$:** For $$h > 0$$, $$0+h \geq 0$$, so $$f(0+h) = 2(0+h) - 1 = 2h - 1$$ 5. **Compute the difference quotient:** $$\frac{f(0+h) - f(0)}{h} = \frac{(2h - 1) - (-1)}{h} = \frac{2h - 1 + 1}{h} = \frac{2h}{h} = 2$$ 6. **Take the limit as $$h \to 0^+$$:** $$f'_+(0) = \lim_{h \to 0^+} 2 = 2$$ **Final answer:** The right-hand derivative of $$f$$ at $$x=0$$ is $$2$$, which corresponds to option (C).