how would you limit the domain to make this function one-to-one and still have the same range? f(x) = 3 - x²…

Explanation:

Step1: Analyze the function type

The function ( f(x) = 3 - x^2 ) is a quadratic function, and its graph is a parabola. The general form of a quadratic function is ( y = ax^2+bx + c ), here ( a=- 1), ( b = 0), ( c = 3). Since ( a<0), the parabola opens downward.

Step2: Find the vertex of the parabola

For a quadratic function ( y=ax^{2}+bx + c), the x - coordinate of the vertex is given by ( x=-\frac{b}{2a}). Substituting ( a=-1) and ( b = 0) into the formula, we get ( x =-\frac{0}{2\times(-1)}=0). The vertex of the parabola ( f(x)=3 - x^{2}) is at ( (0,3) ), which is the maximum point of the parabola (because the parabola opens downward).

Step3: Determine the domain for one - to - one function

A function is one - to - one if no two different inputs give the same output. For a parabola that opens downward, if we consider the domain ( x\geq0), the function is decreasing on this interval (as we move from ( x = 0) to the right, the value of ( y=3 - x^{2}) decreases). In this interval, each ( x) value will map to a unique ( y) value, and the range of the function ( f(x)=3 - x^{2}) for the domain ( x\geq0) is the same as the range of the original function (the range of ( f(x)=3 - x^{2}) is ( (-\infty,3]), and when ( x\geq0), as ( x) varies from ( 0) to (+\infty), ( y) varies from ( 3) to ( -\infty), so the range is still ( (-\infty,3])). If we take the domain ( x\leq0), the function is increasing, but the problem asks for the domain in the form ( x\geq[?]), so we choose the right - hand side of the vertex.

Answer:

( 0 )