(2) If \( \mathrm{f}(x)=\left\{\begin{array}{l}x^{2}-2 x, x \geq 1 \\ 3 x-4, x<1\end{array}\right. \) Discus the differentiability of \( \mathrm{f}(x) \) at \( \mathrm{x}=1 \)

1. **Determine the continuity at \(x=1\):** - For \(x \geq 1\), \[ f(x) = x^2 - 2x. \] Evaluating at \(x = 1\): \[ f(1) = 1^2 - 2(1) = 1 - 2 = -1. \] - For \(x < 1\), \[ f(x) = 3x - 4. \] The left-hand limit as \(x \to 1^{-}\): \[ \lim_{x \to 1^-} f(x) = 3(1) - 4 = 3 - 4 = -1. \] Since the left-hand limit and the value at \(x=1\) are the same, \(f(x)\) is continuous at \(x=1\). 2. **Check differentiability at \(x=1\):** A function is differentiable at a point if the left-hand derivative and the right-hand derivative at that point are equal. - **Right-hand derivative (\(x \geq 1\)):** For \(x \geq 1\), \[ f(x) = x^2 - 2x. \] Differentiate with respect to \(x\): \[ f'(x) = 2x - 2. \] At \(x = 1\): \[ f'(1^+) = 2(1) - 2 = 0. \] - **Left-hand derivative (\(x < 1\)):** For \(x < 1\), \[ f(x) = 3x - 4. \] Differentiate with respect to \(x\): \[ f'(x) = 3. \] At \(x = 1\): \[ f'(1^-) = 3. \] Since the right-hand derivative \(f'(1^+)=0\) and left-hand derivative \(f'(1^-)=3\) are not equal, \(f(x)\) is not differentiable at \(x=1\). 3. **Conclusion:** Although \(f(x)\) is continuous at \(x=1\), it is not differentiable at \(x=1\) because the derivatives from the left and the right do not match.