Question

how would you limit the domain to make this function one-to-one and still have the same range?
f(x) = (x + 3)^2
x ≥ ?

Explanation:

Step1: Analyze the function type

The function \( f(x)=(x + 3)^2 \) is a quadratic function, and its graph is a parabola opening upwards. The vertex of the parabola is at \( x=-3 \) (since for \( y = a(x - h)^2+k \), the vertex is at \( (h,k) \), here \( h=-3,k = 0 \)).

Step2: Determine the one - to - one domain

A parabola is symmetric about its vertex. For a parabola opening upwards, to make it one - to - one (pass the horizontal line test), we can take the domain as \( x\geq \) the x - coordinate of the vertex or \( x\leq \) the x - coordinate of the vertex. The range of \( f(x)=(x + 3)^2 \) is \( y\geq0 \). If we want to keep the same range and make the function one - to - one, we take the right - hand side of the vertex (where the function is increasing). The x - coordinate of the vertex is \( - 3 \), so we limit the domain to \( x\geq - 3 \).

Answer:

\(-3\)