Q.Find all the points of discontinuity of $f$ defined by $f(x) = |x| - |x+1|$.

The key idea is that $f$ is built from absolute value functions, which change expression at the points where their arguments are zero.

We rewrite $|x|$ and $|x+1|$ piecewise:

• $|x| = \begin{cases} -x, & x < 0 \\ x, & x \ge 0 \end{cases}$
• $|x+1| = \begin{cases} -(x+1), & x < -1 \\ x+1, & x \ge -1 \end{cases}$

Now combine over the three intervals determined by $-1$ and $0$:

• For $x < -1$: $f(x) = (-x) - (-(x+1)) = -x + x + 1 = 1$
• For $-1 \le x < 0$: $f(x) = (-x) - (x+1) = -x - x - 1 = -2x - 1$
• For $x \ge 0$: $f(x) = x - (x+1) = -1$

At $x = -1$: left limit $= 1$, right limit $= -2(-1)-1 = 1$, and $f(-1)=1$ — continuous. …

Why This Approach Works

The absolute value function $|t|$ is defined as:

So $|x|$ changes its definition at $x = 0$, and $|x+1|$ changes at $x = -1$. The function $f(x)$ is built from these two pieces. The only places where $f$ could possibly be discontinuous are at these "breakpoints" — $x = -1$ and $x = 0$ — because everywhere else, $f$ is a linear combination of linear functions, hence continuous.

We will check continuity at each breakpoint by computing the left-hand limit, right-hand limit, and the function value. If they all match, the function is continuous there.

Step-by-Step Solution

1. Define the piecewise form of $f(x)$

We need to consider three intervals determined by the breakpoints $-1$ and $0$:

• Interval I: $x < -1$ Here $x < 0$, so $|x| = -x$. Also $x+1 < 0$, so $|x+1| = -(x+1) = -x - 1$. Thus:

• Interval II: $-1 \le x < 0$ Here $x < 0$, so $|x| = -x$. But $x+1 \ge 0$, so $|x+1| = x+1$. Thus:

• Interval III: $x \ge 0$ Here $x \ge 0$, so $|x| = x$. Also $x+1 > 0$, so $|x+1| = x+1$. Thus:

So the piecewise definition is:

2. Check continuity at $x = -1$

• Left-hand limit (as $x \to -1^-$): For $x < -1$, $f(x) = 1$. So:

• Right-hand limit (as $x \to -1^+$): For $-1 \le x < 0$, $f(x) = -2x - 1$. So:

• Function value at $x = -1$: Since $-1$ falls in the second piece ($-1 \le x < 0$), we use $f(-1) = -2(-1) - 1 = 1$. …

Method: Locating Points of Discontinuity in a Sum or Difference of Absolute Values

For a function built from absolute values, only a handful of specific points are even candidates for discontinuity — everywhere else the function is automatically continuous.

Steps

Step 1: Find every point where an expression inside an absolute value changes sign

For each $|\text{expr}(x)|$ appearing in the function, solve $\text{expr}(x) = 0$ — these are the only $x$-values where the function's formula could genuinely change, and are the sole candidates for discontinuity.

Step 2: Rewrite the function as an explicit piecewise definition

Split the real line into the intervals determined by the candidate points from Step 1, and on each interval replace every absolute value with its sign-consistent expression (positive or negated), simplifying to a single formula per interval.

Step 3: Recognise that each piece is automatically continuous on its own open interval

Since each piece simplifies to a polynomial (or another elementary continuous expression), continuity is guaranteed everywhere except possibly at the boundary points themselves. …

Common Mistakes

Mistake 1: Assuming a change in formula automatically means a discontinuity

Why it's wrong: the function's rule genuinely changes at $x=-1$ and $x=0$, but a change in formula does not by itself imply a jump — the two neighbouring pieces can still meet at exactly the same value, exactly as they do here. Correct approach: never conclude discontinuity from the formula changing alone; always compute and compare the actual left-hand limit, right-hand limit, and function value at each breakpoint.

Mistake 2: Checking only one of the two breakpoints …