Q.Show that $f(x) = x^2 + |x - 1|$ is continuous at every real number, using the algebra of continuous functions.

Break $f$ into building blocks whose continuity is already known, then apply the sum rule.

Rather than test each point from scratch, decompose $f$ and apply the algebra rules of §5.

Block 1 — $g(x) = x^2$. This is a polynomial, so by §4 it is continuous at every real number.

Block 2 — $h(x) = |x - 1|$. This is the modulus function $|\cdot|$ (continuous everywhere, §4) composed with the polynomial $x - 1$ (continuous everywhere). By the composite rule of §5, $h$ is continuous at every real number.

Combine with the sum rule. $f = g + h$. Since both $g$ and $h$ are continuous at every real number, the sum rule of §5 (sums of continuous functions are continuous, with no extra condition) gives that $f = g + h$ is continuous at every real number. …

A common mistake is to assume the modulus term $|x-1|$ introduces a discontinuity at $x=1$ because the defining rule changes there. The rule changes, but both pieces meet at the same value, so the function stays continuous. The only genuine restriction in the …