Consider the function y = |x| in the interval [-1, 1]. In this interval, the…

The function y = |x| is continuous on the interval [-1, 1] because there are no breaks or jumps in its graph; specifically, the limit as x approaches 0 equals the function value at 0. However, it is not differentiable at x = 0 because the graph has a sharp corner (cusp) at this point. The left-hand derivative is -1, while the right-hand derivative is +1, meaning they do not match. Since differentiability requires a unique tangent line at every point in the interval, the function fails this condition at x = 0. Therefore, while y = |x| is continuous everywhere in the interval, it lacks differentiability at the origin. This makes Option B the correct choice. Options A and C are incorrect because differentiability implies continuity, so a function cannot be differentiable without being continuous. Option D is wrong since the function is clearly continuous throughout [-1, 1].