Functions: Question 5

Worked solution

Step 1: Find the vertex of the graph

$f(x) = (x+1)^2 - 4$ is an upward-opening parabola with vertex at $(-1,-4)$, the minimum point of the graph.

Step 2: Understand why the vertex splits the domain

A parabola is not one-one over all of $\mathbb{R}$, because every output value (except the minimum itself) is repeated on either side of the vertex. For example:

$f(-3) = (-3+1)^2-4 = 4-4=0 \qquad \text{and} \qquad f(1) = (1+1)^2-4=4-4=0$

Two different inputs, $-3$ and $1$, give the same output, $0$, so $f$ fails the one-one test on $\mathbb{R}$.

To make $f$ one-one, the domain must be restricted to one side only of the vertex $x=-1$:

• $x \ge -1$ (the increasing branch), or
• $x \le -1$ (the decreasing branch)

$f$ is strictly monotonic on each of these branches, so each gives a one-one function.

Step 3: Apply the requirement that the domain contains $x=0$

We need the domain to include $x=0$. Since $0 \ge -1$, the branch $x\ge-1$ contains $x=0$; the branch $x\le-1$ does not.

Step 4: Find the largest such domain

Any subset of $x\ge-1$ that still contains $x=0$ (such as $x\ge0$) would also be one-one, but the question asks for the largest possible domain. The largest one-one domain containing $x=0$ is therefore the full branch: $x \ge -1$

Why the other options are wrong

• B ($x\le-1$): one-one, but excludes $x=0$.
• C ($x\ge0$): one-one and contains $x=0$, but it’s a proper subset of $x\ge-1$, not the largest.
• D ($x\in\mathbb{R}$): the full real line, on which $f$ is not one-one at all.

Final answer

$\boxed{x \ge -1} \quad \text{(Option A)}$