1. **Stating the problem:** We need to check the differentiability of the function $$f(x) = 1 - 3\sqrt{(x-1)^2}$$. 2. **Rewrite the function:** Note that $$\sqrt{(x-1)^2} = |x-1|$$, so the function becomes $$f(x) = 1 - 3|x-1|$$. 3. **Understanding differentiability:** The absolute value function $$|x-1|$$ is differentiable everywhere except possibly at $$x=1$$, where the inside of the absolute value is zero. 4. **Check differentiability at $$x=1$$:** - For $$x > 1$$, $$|x-1| = x-1$$, so $$f(x) = 1 - 3(x-1) = 1 - 3x + 3 = 4 - 3x$$. - For $$x < 1$$, $$|x-1| = -(x-1) = 1 - x$$, so $$f(x) = 1 - 3(1 - x) = 1 - 3 + 3x = 3x - 2$$. 5. **Calculate the left-hand derivative at $$x=1$$:** $$f'_-(1) = \lim_{h \to 0^-} \frac{f(1+h) - f(1)}{h} = \lim_{h \to 0^-} \frac{(3(1+h) - 2) - (1 - 3|1-1|)}{h} = \lim_{h \to 0^-} \frac{3 + 3h - 2 - 1}{h} = \lim_{h \to 0^-} \frac{3h}{h} = 3$$. 6. **Calculate the right-hand derivative at $$x=1$$:** $$f'_+(1) = \lim_{h \to 0^+} \frac{f(1+h) - f(1)}{h} = \lim_{h \to 0^+} \frac{(4 - 3(1+h)) - (1 - 3|1-1|)}{h} = \lim_{h \to 0^+} \frac{4 - 3 - 3h - 1}{h} = \lim_{h \to 0^+} \frac{-3h}{h} = -3$$. 7. **Conclusion:** Since $$f'_-(1) = 3 \neq -3 = f'_+(1)$$, the derivative does not exist at $$x=1$$. 8. **Differentiability elsewhere:** For $$x \neq 1$$, the function is differentiable because it is linear on each side. **Final answer:** The function $$f(x) = 1 - 3\sqrt{(x-1)^2}$$ is differentiable everywhere except at $$x=1$$ where it is not differentiable.