Question

how would you limit the domain to make this function one-to-one and still have the same range? f(x) = x² - 7 x ≥ ?

Explanation:

Step1: Recall the properties of the parabola

The function \( f(x) = x^2 - 7 \) is a parabola that opens upwards (since the coefficient of \( x^2 \) is positive). The vertex of the parabola \( y = ax^2 + bx + c \) is at \( x = -\frac{b}{2a} \). For \( f(x)=x^2 - 7 \), \( a = 1 \), \( b = 0 \), so the vertex is at \( x = 0 \), and the vertex point is \( (0, -7) \).

Step2: Determine the axis of symmetry and one - to - one condition

A parabola is symmetric about its axis of symmetry. For \( f(x)=x^2 - 7 \), the axis of symmetry is \( x = 0 \). A function is one - to - one if it passes the horizontal line test. For a parabola opening upwards, if we take the domain as \( x\geq h \) (where \( h \) is the x - coordinate of the vertex) or \( x\leq h \), the function will be one - to - one. Since we want to keep the same range (the range of \( f(x)=x^2 - 7 \) is \( y\geq - 7 \), because the minimum value of \( x^2 \) is 0, so \( f(x)=x^2 - 7\geq - 7 \)), and we are looking for the domain \( x\geq[\?] \), we take the right - hand side of the vertex (the part of the parabola where \( x\geq0 \)). In this region, as \( x \) increases, \( f(x) \) increases, and it will pass the horizontal line test (be one - to - one) and still have the same range (because the minimum value at \( x = 0 \) is \( - 7 \) and as \( x\) increases, \( f(x)\) can take all values greater than or equal to \( - 7 \)).

Answer:

\( 0 \)