Q.Let $f : [2, \infty) \to \mathbb{R}$ be the function defined by $f(x) = x^2 - 4x + 5$, then the range of $f$ is
(A) $\mathbb{R}$
(B) $[1, \infty)$
(C) $[4, \infty)$
(D) $[5, \infty)$

Concept: Quadratic Range — For a quadratic $f(x) = ax^2 + bx + c$ with $a > 0$, the minimum occurs at $x = -\frac{b}{2a}$; the range is $[f(\text{vertex}), \infty)$.

1. Here $f(x) = x^2 - 4x + 5$. Vertex at $x = -\frac{-4}{2 \cdot 1} = 2$.
2. Since the domain is $[2, \infty)$, the vertex lies at the left endpoint. …

The key here is to recognize that the domain is restricted to $[2, \infty)$, not all real numbers. A common mistake is to find the vertex of the parabola and assume that gives the minimum — but here the vertex lies exactly at the left boundary of the domain, so it's still the minimum, just not for the usual reason.

Let’s break it down.

1. Rewrite the quadratic in vertex form.

   $f(x) = x^2 - 4x + 5$ can be completed as:

   $f(x) = (x^2 - 4x + 4) + 1 = (x - 2)^2 + 1$.

   This tells us the parabola has its vertex at $(2, 1)$ and opens upward (coefficient of $x^2$ is positive).

2. Check where the vertex lies relative to the domain.

   The domain is $[2, \infty)$. The vertex is at $x = 2$, which is included. So the minimum value of $f$ on this domain is $f(2) = 1$.

3. What happens as $x$ increases?

   For $x > 2$, $(x - 2)^2$ grows without bound, so $f(x) \to \infty$. There is no upper limit.

4. Is the function continuous and strictly increasing on $[2, \infty)$? …

Method: Range of a Quadratic on a Restricted Domain

Use this to find the set of outputs of a quadratic when the domain is an interval, not all of $\mathbb{R}$.

Steps

Step 1: Put the quadratic in vertex form.

Complete the square: $ax^2+bx+c=a(x-h)^2+k$, so the vertex sits at $x=h=-\dfrac{b}{2a}$ with value $k$.

Step 2: Locate the vertex relative to the given domain.

If the vertex lies inside the domain, its value is the extreme output. If it lies outside, the function is monotonic on the domain, so the extremes occur at the endpoints. …

Common Mistakes

Mistake 1: Using the whole-line range and ignoring the domain restriction.

Why it's wrong: on a restricted domain the minimum may occur at an endpoint rather than the parabola's vertex. Correct approach: check where the vertex sits relative to the interval, then use the vertex or the endpoint accordingly.

Mistake 2: Reading off the vertex value from the wrong coordinate. …