Q.Find all points of discontinuity of $f$, where $f$ is defined by $f(x) = \begin{cases} |x|+3, & \text{if } x \leq -3 \\ -2x, & \text{if } -3 < x < 3 \\ 6x+2, & \text{if } x \geq 3 \end{cases}$

The key idea is to check continuity at the two potential breakpoints, $x = -3$ and $x = 3$, by comparing left-hand limits, right-hand limits, and the function value at each point.

Step 1: At $x = -3$

• $f(-3) = |-3| + 3 = 3 + 3 = 6$
• Left-hand limit: $\lim_{x \to -3^-} f(x) = \lim_{x \to -3^-} (|x|+3) = 6$
• Right-hand limit: $\lim_{x \to -3^+} f(x) = \lim_{x \to -3^+} (-2x) = -2(-3) = 6$ All three are equal, so $f$ is continuous at $x = -3$.

Step 2: At $x = 3$

• $f(3) = 6(3) + 2 = 18 + 2 = 20$
• Left-hand limit: $\lim_{x \to 3^-} f(x) = \lim_{x \to 3^-} (-2x) = -6$
• Right-hand limit: $\lim_{x \to 3^+} f(x) = \lim_{x \to 3^+} (6x+2) = 20$ …

We need to discuss continuity of a piecewise function. The key idea: a piecewise function can only be discontinuous at the points where the definition changes — here, at $x=-3$ and $x=3$. Everywhere else, each piece is a polynomial (or absolute value, which is also continuous), so the function is automatically continuous on the open intervals $(-\infty, -3)$, $(-3, 3)$, and $(3, \infty)$. Our job is to check what happens at the boundaries.

Let’s go step by step.

1. Check continuity at $x = -3$ At this point, the function uses the first piece: $f(-3) = |{-3}| + 3 = 3 + 3 = 6$. Now find the left-hand limit as $x \to -3^-$ (from values less than $-3$). For $x \leq -3$, the rule is $|x|+3$. Since $x$ is negative, $|x| = -x$, so $|x|+3 = -x+3$.

For the right-hand limit as $x \to -3^+$ (from values greater than $-3$), we use the second piece: $f(x) = -2x$ for $-3 < x < 3$.

All three values — left limit, right limit, and function value — are equal to $6$. So $f$ is continuous at $x = -3$.

2. Check continuity at $x = 3$ At $x = 3$, the function uses the third piece: $f(3) = 6(3)+2 = 18+2 = 20$. Left-hand limit as $x \to 3^-$ uses the second piece:

Right-hand limit as $x \to 3^+$ uses the third piece: …

Method: Finding Discontinuities of a Multi-Piece Function Involving an Absolute Value

With three (or more) pieces, there are multiple seam points to check — and when one piece contains $|x|$, that absolute value must first be rewritten as a signed expression valid on the relevant side of the seam before any limit is taken.

Steps

Step 1: List every seam value

Identify every $x$-value where the definition of $f$ changes (here, $x=-3$ and $x=3$) — these are the only candidates for discontinuity.

Step 2: Rewrite any absolute-value piece using the correct sign for the region near each seam

For $x\le-3$ (so $x$ is negative), $|x|=-x$; near a seam where $x$ could be positive, use $|x|=x$ instead. Get this sign right before taking any limit.

Step 3: At each seam, compute the function value and both one-sided limits

using the piece (with the correctly signed absolute value, where relevant) that governs each side. …

Common Mistakes

Mistake 1: Using the wrong sign when simplifying $|x|$ near $x=-3$

Why it's wrong: for $x\le-3$, $x$ is negative, so $|x|=-x$, not $x$ — using $|x|=x$ by default (a common reflex) flips the sign of the whole expression and gives a wrong limit value. Correct approach: always determine the sign of $x$ in the relevant region first, then substitute $|x|=x$ or $|x|=-x$ accordingly before taking the limit.

Mistake 2: Assuming that because one seam ($x=-3$) turned out continuous, the other seam ($x=3$) must be too …