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Sagot :
To find the inverse of the function [tex]\( m(x) = x^2 - 17x \)[/tex], we want to solve for [tex]\( x \)[/tex] in terms of [tex]\( y \)[/tex], where [tex]\( y = m(x) \)[/tex].
1. Start by setting [tex]\( y = x^2 - 17x \)[/tex].
[tex]\[ y = x^2 - 17x \][/tex]
2. Rearrange the equation to solve for [tex]\( x \)[/tex].
[tex]\[ x^2 - 17x - y = 0 \][/tex]
This is a quadratic equation in standard form [tex]\( ax^2 + bx + c = 0 \)[/tex], where [tex]\( a = 1 \)[/tex], [tex]\( b = -17 \)[/tex], and [tex]\( c = -y \)[/tex].
3. Solve the quadratic equation using the quadratic formula:
[tex]\[ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \][/tex]
Plug in the values of [tex]\( a \)[/tex], [tex]\( b \)[/tex], and [tex]\( c \)[/tex]:
[tex]\[ x = \frac{17 \pm \sqrt{17^2 - 4 \cdot 1 \cdot (-y)}}{2 \cdot 1} \][/tex]
[tex]\[ x = \frac{17 \pm \sqrt{289 + 4y}}{2} \][/tex]
[tex]\[ x = \frac{17 \pm \sqrt{4y + 289}}{2} \][/tex]
Therefore, we have two solutions:
[tex]\[ x_1 = \frac{17 + \sqrt{4y + 289}}{2} \][/tex]
[tex]\[ x_2 = \frac{17 - \sqrt{4y + 289}}{2} \][/tex]
4. To determine which expression to use for the inverse function [tex]\( m^{-1}(x) \)[/tex], we need to consider the domain restriction. Given the domain restriction [tex]\( x \geq \frac{17}{2} \)[/tex]:
[tex]\[ \frac{17}{2} = 8.5 \][/tex]
5. Substitute [tex]\( x = 8.5 \)[/tex] into both solutions to check which one satisfies the domain restriction:
[tex]\[ x_1 = \frac{17 + \sqrt{4y + 289}}{2} = 17 - \frac{\sqrt{4y + 289}}{2} \][/tex]
Given the restriction [tex]\( x \geq 8.5 \)[/tex]:
[tex]\[ \frac{17 - \sqrt{4y + 289}}{2} \leq 8.5 \][/tex]
Since [tex]\( 17 - \sqrt{4y + 289} \geq 8.5 \)[/tex], the feasible solution satisfying the domain restriction is:
[tex]\[ x_2 = \frac{17 + \sqrt{4y + 289}}{2} \][/tex]
6. Therefore, considering the inverse function [tex]\( m^{-1}(x) \)[/tex] with the given domain restriction:
[tex]\[ m^{-1}(x) = \frac{17 - \sqrt{4y + 289}}{2} \][/tex]
Thus, the correct statement is:
The domain restriction [tex]\( x \geq \frac{17}{2} \)[/tex] results in [tex]\( m^{-1}(x) = \frac{17 - \sqrt{x + \frac{289}{4}}} \)[/tex].
1. Start by setting [tex]\( y = x^2 - 17x \)[/tex].
[tex]\[ y = x^2 - 17x \][/tex]
2. Rearrange the equation to solve for [tex]\( x \)[/tex].
[tex]\[ x^2 - 17x - y = 0 \][/tex]
This is a quadratic equation in standard form [tex]\( ax^2 + bx + c = 0 \)[/tex], where [tex]\( a = 1 \)[/tex], [tex]\( b = -17 \)[/tex], and [tex]\( c = -y \)[/tex].
3. Solve the quadratic equation using the quadratic formula:
[tex]\[ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \][/tex]
Plug in the values of [tex]\( a \)[/tex], [tex]\( b \)[/tex], and [tex]\( c \)[/tex]:
[tex]\[ x = \frac{17 \pm \sqrt{17^2 - 4 \cdot 1 \cdot (-y)}}{2 \cdot 1} \][/tex]
[tex]\[ x = \frac{17 \pm \sqrt{289 + 4y}}{2} \][/tex]
[tex]\[ x = \frac{17 \pm \sqrt{4y + 289}}{2} \][/tex]
Therefore, we have two solutions:
[tex]\[ x_1 = \frac{17 + \sqrt{4y + 289}}{2} \][/tex]
[tex]\[ x_2 = \frac{17 - \sqrt{4y + 289}}{2} \][/tex]
4. To determine which expression to use for the inverse function [tex]\( m^{-1}(x) \)[/tex], we need to consider the domain restriction. Given the domain restriction [tex]\( x \geq \frac{17}{2} \)[/tex]:
[tex]\[ \frac{17}{2} = 8.5 \][/tex]
5. Substitute [tex]\( x = 8.5 \)[/tex] into both solutions to check which one satisfies the domain restriction:
[tex]\[ x_1 = \frac{17 + \sqrt{4y + 289}}{2} = 17 - \frac{\sqrt{4y + 289}}{2} \][/tex]
Given the restriction [tex]\( x \geq 8.5 \)[/tex]:
[tex]\[ \frac{17 - \sqrt{4y + 289}}{2} \leq 8.5 \][/tex]
Since [tex]\( 17 - \sqrt{4y + 289} \geq 8.5 \)[/tex], the feasible solution satisfying the domain restriction is:
[tex]\[ x_2 = \frac{17 + \sqrt{4y + 289}}{2} \][/tex]
6. Therefore, considering the inverse function [tex]\( m^{-1}(x) \)[/tex] with the given domain restriction:
[tex]\[ m^{-1}(x) = \frac{17 - \sqrt{4y + 289}}{2} \][/tex]
Thus, the correct statement is:
The domain restriction [tex]\( x \geq \frac{17}{2} \)[/tex] results in [tex]\( m^{-1}(x) = \frac{17 - \sqrt{x + \frac{289}{4}}} \)[/tex].
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