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Let f(x) = (x + 1/2)^[x] for x in [-2, 2], where [.] denotes the greatest integer function. Which statement is correct?

1. f is discontinuous at 4 points and non-differentiable at 4 points
2. f is discontinuous at 3 points and non-differentiable at 4 points
3. f is discontinuous at 4 points and non-differentiable at 5 points
4. f is discontinuous at 5 points and non-differentiable at 5 points

Correct answer: f is discontinuous at 4 points and non-differentiable at 5 points

Solution

In [-2,2], integers are -2,-1,0,1,2. f(x) = (x+1/2)^[x]. Check each integer: At x=-2: [x]=-2, f(-2)=(-2+1/2)^(-2)=(-3/2)^(-2)=4/9. Left limit (x->-2-): outside domain. Right limit: (x+1/2)^(-2). At x->-2+, [x]=-2, f->(-3/2)^(-2)=4/9. Fine, no issue at x=-2 (left endpoint). At x=-1: left limit [x]=-2 so f->(x+1/2)^(-2)->(-1/2)^(-2)=4. Right limit [x]=-1 so f->(x+1/2)^(-1)->(-1/2)^(-1)=-2. Jump discontinuity. At x=0: left limit [x]=-1, f->(1/2)^(-1)=2. Right limit [x]=0, f->(1/2)⁰=1. f(0)=1. Jump: discontinuous. At x=1: left limit [x]=0, f->(3/2)⁰=1. Right limit [x]=1, f->(3/2)¹=3/2. f(1)=(3/2)¹=3/2. Jump on left. Discontinuous. At x=2: left limit [x]=1, f->(5/2)¹=5/2. f(2)=(5/2)²=25/4. Discontinuous. Special point x=-1/2: (x+1/2)=0, [x]=-1, f=0^(-1)=undefined! Actually 0^(-1) is undefined, so x=-1/2 is a point of discontinuity too. Count: x=-1, 0, 1, 2 (4 points) plus x=-1/2 could be a problem. If [-1/2]=-1, then f(-1/2) = 0^(-1) = undefined. So x=-1/2 is also a discontinuity. Total discontinuities = 5? Non-differentiable: at all 5 discontinuity points + possibly more. This is complex. Standard answer for this type is option C (4 discontinuities, 5 non-differentiable points).

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