Q.Find the maximum and the minimum values, if any, of the function $f$ given by $f(x) = x^2,\ x \in \mathbb{R}$.

Concept: Quadratic function — vertex form.

Since $f(x) = x^2$ is a parabola opening upwards with vertex at $(0,0)$, the minimum occurs at the vertex.

• For any real $x$, $x^2 \ge 0$, so the smallest value is $0$ at $x = 0$.
• As $x \to \pm\infty$, $x^2 \to \infty$, so there is no maximum value — the function is unbounded above.

The Mean Value Theorem (MVT) is a powerful tool for analyzing function behaviour, but here it’s not needed — the shape of $x^2$ is simpler. The key is to think about what “maximum” and “minimum” mean for a function defined on the entire real line. A minimum is the smallest output the function ever takes; a maximum is the largest. For $x^2$, the graph is a parabola opening upward, with its vertex at the origin. That vertex is clearly the lowest point. But does the parabola have a highest point? No — as you move farther from zero in either direction, the squares get larger without any bound.

Let’s walk through the reasoning step by step.

1. Understand the domain and range.

   The domain is all real numbers $\mathbb{R}$. The output $x^2$ is always non-negative: $x^2 \ge 0$ for every $x \in \mathbb{R}$. So the range is $[0, \infty)$.

2. Check for a minimum.

   Since $x^2 \ge 0$, the smallest possible value is $0$. Does the function actually achieve $0$? Yes — at $x = 0$, we have $f(0) = 0^2 = 0$. So $0$ is the global minimum of $f$ on $\mathbb{R}$.

3. Check for a maximum.

   Is there a largest value? Suppose someone claims $M$ is the maximum. Then for any $x$, we must have $x^2 \le M$. But pick $x = \sqrt{M} + 1$; then $x^2 = M + 2\sqrt{M} + 1 > M$, contradicting the claim. No matter how large $M$ is, you can always find an $x$ whose square exceeds it. Hence, no maximum exists.

4. Formal justification using limits. We can also argue: $\lim_{x \to \pm\infty} x^2 = \infty$, so the function is unbounded above. A maximum requires an upper bound that is actually attained; here, no such bound exists.

Method: Deciding Whether a Function Has a Global Maximum/Minimum Over All of $\mathbb{R}$

This method applies to "find the maximum and minimum values, if any" questions where the domain is the whole real line, not a closed interval — the "if any" is a hint that one or both extrema might not exist.

Steps

Step 1: Determine the overall behaviour of the function as $x \to \pm\infty$.

Does the function grow without bound in either direction, or does it stay bounded?

Step 2: If the function is bounded on one side, check whether that bound is actually attained by some $x$ in the domain.

If it is, that value is the global extremum on that side; if it's only an approached-but-never-reached bound, there is no extremum there.

Step 3: If the function is unbounded in a direction, conclude "no maximum" (or "no minimum") on that side — never write "the maximum is $\infty$."

Infinity is not a value the function actually takes, so it cannot be a maximum or minimum in the mathematical sense.

Step 4: For a quadratic specifically, use the vertex to locate the one-sided extremum quickly.

For $f(x) = ax^2+bx+c$, the vertex $x = -\dfrac{b}{2a}$ gives the minimum (if $a>0$) or the maximum (if $a<0$); the opposite extremum never exists on $\mathbb{R}$, since a parabola is unbounded on the other side.

Common Mistakes

Mistake 1: Writing "the maximum value is $\infty$" instead of stating that no maximum exists.

Why it's wrong: $\infty$ is not a real number the function actually attains, so calling it "the maximum" is mathematically incorrect — the precise statement is that no maximum value exists because the function is unbounded above, which is a different claim from assigning it a symbolic value. Correct approach: explicitly write "no maximum value exists" rather than treating $\infty$ as an attained output.