Q.Show that the function $f$ defined by $f(x) = |1 - x + |x||$, where $x$ is any real number, is a continuous function.

Concept: Continuity At A Point — a function is continuous at $x = a$ if $\lim_{x \to a} f(x) = f(a)$. We check this for all real $x$.

Step 1: Simplify the inner expression.

Recall $|x| = x$ for $x \ge 0$ and $|x| = -x$ for $x < 0$.

So $1 - x + |x|$ becomes:

• For $x \ge 0$: $1 - x + x = 1$
• For $x < 0$: $1 - x - x = 1 - 2x$

Step 2: Write $f(x)$ piecewise.

For $x < 0$, $1 - 2x > 0$ (since $x<0$ gives $-2x>0$), so $|1-2x| = 1-2x$.

Thus:

Step 3: Check continuity at the only potential trouble point, $x=0$. …

We need to show that $f(x) = |1 - x + |x||$ is continuous for all real $x$. The function involves an absolute value inside another absolute value. The standard way to handle such nested absolute values is to remove them by considering the cases where the inner expression changes sign.

The innermost absolute value is $|x|$, which changes behaviour at $x = 0$. So we split the domain into $x \geq 0$ and $x < 0$.

1. Case 1: $x \geq 0$ Here $|x| = x$. Substitute into $f$:

So for all $x \geq 0$, $f(x) = 1$, a constant function. Constant functions are continuous everywhere on their domain.

2. Case 2: $x < 0$ Here $|x| = -x$. Substitute:

Now we have $|1 - 2x|$. This absolute value changes sign when $1 - 2x = 0$, i.e., $x = \frac{1}{2}$. But note: we are in the region $x < 0$, and $\frac{1}{2} > 0$, so the point $x = \frac{1}{2}$ is not in this region. Therefore, for all $x < 0$, the expression $1 - 2x$ is always positive (since $x$ is negative, $-2x$ is positive, so $1 - 2x > 1 > 0$). Hence:

So for $x < 0$, $f(x) = 1 - 2x$, a linear polynomial. Linear functions are continuous everywhere on their domain.

3. Check continuity at the boundary $x = 0$ The function is defined piecewise:

At $x = 0$, we compute: …

Method: Simplifying a Nested Absolute Value by Case-Splitting on Sign, Then Testing Continuity

This method applies whenever a function contains one or more absolute values, and you need to first remove them (by splitting into cases) before the piece can be analysed for continuity.

Steps

Step 1: Work from the innermost absolute value outward.

Identify the expression inside the innermost $|\cdot|$ and find where it changes sign — for $|x|$, that's at $x=0$. Split the real line into the regions where that inner expression is $\ge 0$ and $<0$.

Step 2: Substitute the correct sign rule into the outer expression, one region at a time.

For each region, replace the inner absolute value with $+(\text{expression})$ or $-(\text{expression})$ as appropriate, and simplify what remains inside any outer absolute value signs.

Step 3: Within each region, check whether the new, simplified inner quantity changes sign inside that region. …

Common Mistakes

Mistake 1: Removing the outer absolute value bars using the sign of the original variable, instead of the sign of the simplified expression actually inside those bars.

Why it's wrong: after substituting the inner absolute value's sign rule, the expression inside the outer bars is a different expression (e.g. $1-2x$, not $x$ itself) — its own sign, in the current region, is what determines how to remove the outer bars, not the sign of $x$. Correct approach: after each substitution, explicitly re-check the sign of the new expression within that region before removing the next layer of absolute value.

Mistake 2: Assuming the newly-simplified expression's zero must fall inside the current region.

Why it's wrong: a sign-change point found by solving (e.g.) $1-2x=0$ can land outside the region under consideration ($x=\tfrac12$ is not in $x<0$) — treating it as a further internal split point when it isn't creates a false extra "piece" that doesn't actually exist. Correct approach: always check whether the computed zero actually lies within the current region before splitting further. …