Q.State whether True or False: If $f$ is continuous on its domain $D$, then $|f|$ is also continuous on $D$.

Concept: Continuity At A Point — The absolute value function is continuous everywhere, and the composition of continuous functions is continuous.

1. For any $c \in D$, $f$ is continuous at $c$, so $\lim_{x \to c} f(x) = f(c)$.
2. The function $g(t) = |t|$ is continuous at every real $t$. …

The heart of this question is not about some special property of $f$ — it's about how continuity behaves under composition. When you see $|f|$, think: "first apply $f$, then apply the absolute value function." If both steps preserve continuity, the result must be continuous.

Let's unpack why.

1. Recall the definition of continuity at a point.

   A function $g$ is continuous at $x = a$ if $\lim_{x \to a} g(x) = g(a)$. For $|f|$ to be continuous at any $a$ in $D$, we need $\lim_{x \to a} |f(x)| = |f(a)|$.

2. The absolute value function itself is continuous everywhere.

   This is a standard result: $h(t) = |t|$ is continuous for all real $t$. You can prove it quickly: for any $c$, $\lim_{t \to c} |t| = |c|$, because $|t| - |c| \leq |t - c|$ (reverse triangle inequality), so as $t \to c$, $|t| \to |c|$.

3. Now use the composition theorem for continuity.

   A key theorem says: if $g$ is continuous at $a$ and $h$ is continuous at $g(a)$, then $h \circ g$ is continuous at $a$. Here, $g = f$ and $h = |\cdot|$. Since $f$ is continuous at every $a \in D$ (by assumption), and $|\cdot|$ is continuous at $f(a)$, the composition $|f|$ is continuous at every $a \in D$.

4. A common worry: what if $f$ is not differentiable or has sharp corners? That doesn't matter. Continuity is a weaker condition than differentiability. Even if $f$ has a cusp or a corner, $|f|$ can still be continuous. For example, $f(x) = x$ is continuous, and $|x|$ is continuous (though not differentiable at $0$). The composition theorem only cares about continuity, not smoothness. …

Method: Proving Continuity of $|f(x)|$ Using the Composition Rule

This method solves "is $|f|$ continuous, given $f$ is continuous" reasoning questions, by recognising $|f(x)|$ as a composition of two continuous functions.

Steps

Step 1: Recognise the composition

Write $|f(x)|$ as $(h\circ f)(x)=h\big(f(x)\big)$, where $h(t)=|t|$.

Step 2: Recall that the outer function is continuous everywhere

$h(t)=|t|$ is continuous at every real number $t$ — a standard fact.

Step 3: Apply the composition-of-continuous-functions rule

If $f$ is continuous at $a$, and $h$ is continuous at the value $f(a)$, then the composite $h\circ f$ is continuous at $a$. …

Common Mistakes

Mistake 1: False analogy with the differentiability counterexample

Because $|x|$ is the famous example of a function that is continuous but NOT differentiable at a corner, students sometimes assume $|f|$ must similarly "break" continuity wherever $f(x)=0$. Why it's wrong: this confuses a differentiability issue with a continuity claim — the corner in $|x|$ at $0$ is still perfectly continuous, it just lacks a single tangent line. Correct approach: keep the two properties separate; the question only asks about continuity, which the corner does not disturb.

Mistake 2: Proving it with one example instead of the general rule …