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

Explanation:

Step1: Analyze the function's graph

The function ( f(x) = -x^2 - 1 ) is a parabola opening downward (since the coefficient of ( x^2 ) is negative). The vertex of the parabola ( y = ax^2 + bx + c ) is at ( x = -\frac{b}{2a} ). For ( f(x) = -x^2 - 1 ), ( a = -1 ), ( b = 0 ), so the vertex is at ( x = 0 ), and the vertex point is ( (0, -1) ) (since ( f(0)= -0 - 1=-1 )).

Step2: Determine the domain for one - to - one

A function is one - to - one if it passes the horizontal line test. For a downward - opening parabola, to make it one - to - one and keep the same range, we can take the right - hand side of the vertex (where ( x\geq0 )) or the left - hand side (where ( x\leq0 )). Looking at the graph, the vertex is at ( x = 0 ). If we want ( x\geq[?] ) to make the function one - to - one and have the same range, we take ( x\geq0 ) because the right half of the parabola (for ( x\geq0 )) is decreasing, and it will pass the horizontal line test.

Answer:

( 0 )