Q.Show that the function defined by $f(x) = |\cos x|$ is a continuous function.

The key idea is that continuity is preserved under composition: $|\cdot|$ is continuous everywhere, and $\cos x$ is continuous everywhere, so their composition $|\cos x|$ is continuous.

Step 1: The absolute value function $g(t) = |t|$ is continuous for all $t \in \mathbb{R}$ (it is a basic continuous function).

Step 2: The cosine function $h(x) = \cos x$ is continuous for all $x \in \mathbb{R}$ (standard trigonometric continuity). …

Why This Approach Works

The key insight here is that we don't need to wrestle with $\epsilon$-$\delta$ proofs or check continuity at every single point manually. Instead, we can use a powerful theorem: the composition of continuous functions is continuous.

Think of $f(x) = |\cos x|$ as two machines working in sequence:

1. First, the "cosine machine" takes $x$ and produces $\cos x$.
2. Then, the "absolute value machine" takes that result and produces $|\cos x|$.

If each machine individually produces continuous outputs, then the combined machine also produces continuous outputs. This is the composition rule in action.

Step-by-Step Solution

1. Identify the two functions being composed.

We have $f(x) = |\cos x|$. Let:

• $g(x) = \cos x$ (the inner function)
• $h(t) = |t|$ (the outer function)

Then $f(x) = h(g(x)) = |\cos x|$.

2. Show that $g(x) = \cos x$ is continuous everywhere.

The cosine function is continuous for all real numbers. This is a standard result from trigonometry — its graph is a smooth, unbroken wave with no jumps, holes, or vertical asymptotes. Formally, for any real $a$:

3. Show that $h(t) = |t|$ is continuous everywhere.

The absolute value function is also continuous for all real numbers. For any real $a$:

This is easy to see geometrically — the graph of $|t|$ is a V-shape with no breaks. The only potential worry is at $t = 0$, but even there, the left-hand limit and right-hand limit both equal $0$, which equals $|0|$.

4. Apply the composition theorem.

Since:

• $g(x) = \cos x$ is continuous at every real number $x$
• $h(t) = |t|$ is continuous at every real number $t$, including at $t = \cos x$ (since $\cos x$ always lies between $-1$ and $1$, and $|t|$ is continuous everywhere) …

Method: Composition Proofs Involving an Absolute Value Outer Function

When the outer function is $|\cdot|$, the proof follows the same composition pattern, but it's worth explicitly addressing the corner the absolute value creates, since that's where a student's doubt usually surfaces.

Steps

Step 1: Identify inner and outer functions

Write the given function as $h(g(x))$, with $g(x)$ the expression being fed into an absolute value, and $h(t) = |t|$.

Step 2: Confirm the inner function is continuous everywhere

If $g$ is a standard trigonometric or algebraic function, state the standard result (e.g. sine and cosine are continuous for all real numbers).

Step 3: Confirm the outer function $|t|$ is continuous for every real $t$, including $t = 0$ …

Common Mistakes

Mistake 1: Treating the corner at $\cos x = 0$ as a break in continuity

Why it's wrong: $|\cos x|$ has sharp points wherever $\cos x = 0$ (e.g. $x=\frac{\pi}{2}$) because the absolute value reflects the negative part of cosine upward — but a sharp corner is a differentiability issue, not a continuity issue. The left-hand limit, right-hand limit, and function value all still agree at those points. Correct approach: separately verify the three continuity conditions (or just apply the composition theorem) at a corner point rather than assuming a visible kink implies a jump.

Mistake 2: Forgetting that the composition theorem needs both functions continuous, not just one …