Q.Discuss the continuity of the function $f$ given by $f(x) = x^3 + x^2 - 1$.

Concept: Continuity At A Point — A polynomial is continuous at every real number because its limit equals its value at that point.

Since $f(x) = x^3 + x^2 - 1$ is a polynomial, it is continuous for all $x \in \mathbb{R}$. There is no need to check individual points; the fun …

Why This Problem Is Simpler Than It Looks

Many students, when asked to "discuss the continuity" of a function, immediately reach for the three-part definition: check if $f(a)$ exists, if $\lim_{x\to a}f(x)$ exists, and if they are equal. That is the correct general procedure — but it is overkill here.

The key insight: polynomials are the "nice" functions of calculus. They are built only from powers of $x$ with constant coefficients, using addition and multiplication. No division by zero, no piecewise definitions, no radicals that could go negative, no logarithms or trig functions with restricted domains. A polynomial is defined and smooth everywhere on the real line.

This is a theorem you can rely on in exams. It follows from two simpler facts: the identity function $g(x)=x$ is continuous, and the constant function $h(x)=c$ is continuous. Since sums, products, and constant multiples of continuous functions are continuous, any polynomial — being a finite combination of these — inherits continuity everywhere.

So for $f(x)=x^3+x^2-1$, we already know the answer: it is continuous on $\mathbb{R}$. But let us verify it properly, step by step, so the reasoning is clear.

Step-by-Step Verification

1. Choose an arbitrary point $a \in \mathbb{R}$.

Continuity is a local property — we check it at each point individually. Since the domain is all real numbers, we pick any real $a$ and show continuity there.

2. Check that $f(a)$ is defined.

$f(a) = a^3 + a^2 - 1$. This is a real number for every real $a$. No issues.

3. Compute $\lim_{x \to a} f(x)$.

Because $f$ is a polynomial, the limit as $x$ approaches $a$ is simply $f(a)$. We can justify this using the limit laws:

• $\lim_{x \to a} x = a$ (the identity function is continuous) …

Method: Using the Algebra of Continuous Functions to Shortcut a Continuity Check

This method proves continuity of a "built-up" function — like a polynomial — quickly, by citing the standard theorems about sums, products, and constant multiples of already-known continuous functions, instead of re-deriving the limit from scratch.

Steps

Step 1: Recognise the function as built from simpler continuous pieces

Identify that the function is a finite combination — via addition, subtraction, multiplication, or constant scaling — of functions already known to be continuous everywhere, most commonly the identity function $g(x) = x$ and constant functions.

Step 2: Cite the relevant continuity theorems

Recall that if $g$ and $h$ are continuous at a point (or everywhere), then so are $g + h$, $g - h$, $g \cdot h$, and $k \cdot g$ for a constant $k$ (and $g/h$ wherever $h \neq 0$).

Step 3: Build up the target function piece by piece …

Common Mistakes

Mistake 1: Thinking a turning point or "bump" in the graph signals a discontinuity

Why it's wrong: turning points, local maxima/minima, and changes in curvature are features of a function's shape, not of its continuity — continuity only concerns whether the limit matches the function value at each point, which every polynomial automatically satisfies. Correct approach: recall that all polynomials are continuous everywhere as a standard theorem, and don't confuse "wiggly-looking" with "broken."

Mistake 2: Re-deriving the limit from scratch with an $\varepsilon$–$\delta$ argument when a theorem already applies …