Let's analyze the function f(x) defined piecewise on the interval [-1, 3]. The function involves absolute value |x| and the greatest integer function [x] (also known as the floor function). We need to find the points where f is discontinuous.

The function is defined as:

$f\left(x\right)=\{\begin{array}{cc}\left|x\right|+\left[x\right],\hfill & -1\le x<1\hfill \\ x+\left|x\right|,\hfill & 1\le x<2\hfill \\ x+\left[x\right],\hfill & 2\le x\le 3\hfill \end{array}$

We know that the greatest integer function [x] is discontinuous at all integer points. The absolute value function |x| is continuous everywhere but has a sharp corner (non-differentiable) at x=0. However, our main concern is discontinuity of f(x).

Let's break the domain into subintervals based on the definition of f and the behavior of [x] and |x|.

Step 1: Analyze the first piece: f(x) = |x| + [x] for -1 ≤ x < 1

In this interval, [x] takes constant values on integer intervals:

• For -1 ≤ x < 0: [x] = -1 (since -1 is the greatest integer ≤ x in this range). Also, |x| = -x (since x is negative). So f(x) = -x + (-1) = -x - 1. This is a linear function, continuous.
• For 0 ≤ x < 1: [x] = 0. Also, |x| = x. So f(x) = x + 0 = x. This is linear, continuous.

However, we must check the boundaries:

• At x = -1: f(-1) = | -1 | + [ -1 ] = 1 + (-1) = 0. The left limit at x=-1 is not defined since domain starts at -1. But we check right limit: as x→-1⁺ (from right, i.e., just greater than -1), f(x) = -x -1 → -(-1) -1 = 1-1=0. So continuous at x=-1.
• At x = 0: Left limit (x→0⁻): f(x) = -x -1 → -0 -1 = -1. Right limit (x→0⁺): f(x) = x → 0. f(0) = |0| + [0] = 0 + 0 = 0. Since left limit (-1) ≠ right limit (0) ≠ f(0) (0), there is a discontinuity at x=0.
• At x=1: This piece is defined for x<1 only. So we check the limit as x→1⁻: for x just less than 1, [x]=0, |x|=x, so f(x)=x → 1. But f(1) is defined by the second piece.

So, in the first piece, discontinuity at x=0.

Step 2: Analyze the second piece: f(x) = x + |x| for 1 ≤ x < 2

In this interval, x is positive (since x≥1), so |x| = x. Therefore, f(x) = x + x = 2x. This is linear, continuous.

Check boundaries:

• At x=1: Left limit (from first piece, x→1⁻): f(x)=x → 1. Right limit (x→1⁺): f(x)=2x → 2. f(1)=2*1=2. Since left limit (1) ≠ right limit (2), there is a discontinuity at x=1.
• At x=2: This piece is defined for x<2. So limit as x→2⁻: f(x)=2x → 4. f(2) is defined by the third piece.

So, discontinuity at x=1.

Step 3: Analyze the third piece: f(x) = x + [x] for 2 ≤ x ≤ 3

In this interval:

• For 2 ≤ x < 3: [x] = 2, so f(x) = x + 2. This is linear, continuous.
• For x = 3: [3] = 3, so f(3)=3+3=6.

Check boundaries:

• At x=2: Left limit (from second piece, x→2⁻): f(x)=2x → 4. Right limit (x→2⁺): f(x)=x+2 → 2+2=4. f(2)=2+2=4. So continuous at x=2.
• At x=3: Only left limit exists. As x→3⁻: f(x)=x+2 → 5. But f(3)=6. So 5 ≠ 6, discontinuity at x=3.

So, discontinuity at x=3.

Summary of discontinuities:

• x = 0 (from first piece)
• x = 1 (at the junction of first and second piece)
• x = 3 (at the end of third piece)

Also, note that the greatest integer function [x] is discontinuous at integers, but in our function f, it is combined with other functions. We have already considered the integer points in our domain: x=-1,0,1,2,3. We found that x=-1 and x=2 are continuous. So only three points of discontinuity: x=0, x=1, x=3.

Final Answer: f is discontinuous at only three points.

Related Topics and Formulae

Greatest Integer Function (Floor function) [x]: It returns the greatest integer less than or equal to x. It is discontinuous at all integer points.

Absolute Value Function |x|: Defined as |x| = x if x≥0, and |x| = -x if x<0. It is continuous everywhere but not differentiable at x=0.

Continuity of a Function: A function f is continuous at a point x=a if: $\underset{x\to a}{\lim }f\left(x\right)=f\left(a\right)$. For piecewise functions, check left-hand limit, right-hand limit, and the value at the point.