Answer:
The correct option is;
Step-by-step explanation:
The given information is that m(x) = x² - 17·x
The above equation can be written in the form;
y = x² - 17·x
Therefore;
0 = x² - 17·x - y
From the general solution of a quadratic equation, 0 = a·x² + b·x + c we have;
![x = \dfrac{-b\pm \sqrt{b^{2}-4\cdot a\cdot c}}{2\cdot a}](https://tex.z-dn.net/?f=x%20%3D%20%5Cdfrac%7B-b%5Cpm%20%5Csqrt%7Bb%5E%7B2%7D-4%5Ccdot%20a%5Ccdot%20c%7D%7D%7B2%5Ccdot%20a%7D)
By comparison to the equation,0 = x² - 17·x - y, we have;
a = 1, b = -17, and c = -y
Substituting the values of a, b and c into the formula for the general solution of a quadratic equation, we have;
![x = \dfrac{-(-17)\pm \sqrt{(-17)^{2}-4\times (1) \times (-y)}}{2\times (1)} = \dfrac{17\pm \sqrt{289+4\cdot y}}{2}](https://tex.z-dn.net/?f=x%20%3D%20%5Cdfrac%7B-%28-17%29%5Cpm%20%5Csqrt%7B%28-17%29%5E%7B2%7D-4%5Ctimes%20%281%29%20%5Ctimes%20%28-y%29%7D%7D%7B2%5Ctimes%20%281%29%7D%20%3D%20%5Cdfrac%7B17%5Cpm%20%5Csqrt%7B289%2B4%5Ccdot%20y%7D%7D%7B2%7D)
Which can be simplified as follows;
![x = \dfrac{17\pm \sqrt{289+4\cdot y}}{2}= \dfrac{17}{2} \pm \dfrac{1}{2} \times \sqrt{289+4\cdot y}} = \dfrac{17}{2} \pm \sqrt{\dfrac{289}{4} +\dfrac{4\cdot y}{4} }}](https://tex.z-dn.net/?f=x%20%3D%20%20%5Cdfrac%7B17%5Cpm%20%5Csqrt%7B289%2B4%5Ccdot%20y%7D%7D%7B2%7D%3D%20%5Cdfrac%7B17%7D%7B2%7D%20%5Cpm%20%5Cdfrac%7B1%7D%7B2%7D%20%20%5Ctimes%20%5Csqrt%7B289%2B4%5Ccdot%20y%7D%7D%20%3D%20%5Cdfrac%7B17%7D%7B2%7D%20%5Cpm%20%5Csqrt%7B%5Cdfrac%7B289%7D%7B4%7D%20%2B%5Cdfrac%7B4%5Ccdot%20y%7D%7B4%7D%20%7D%7D)
And further simplified as follows;
![x = \dfrac{17}{2} \pm \sqrt{\dfrac{289}{4} +y }} = \dfrac{17}{2} \pm \sqrt{y + \dfrac{289}{4} }}](https://tex.z-dn.net/?f=x%20%3D%20%5Cdfrac%7B17%7D%7B2%7D%20%5Cpm%20%5Csqrt%7B%5Cdfrac%7B289%7D%7B4%7D%20%2By%20%7D%7D%20%3D%20%5Cdfrac%7B17%7D%7B2%7D%20%5Cpm%20%5Csqrt%7By%20%2B%20%5Cdfrac%7B289%7D%7B4%7D%20%7D%7D)
Interchanging x and y in the function of the inverse, m⁻¹(x), we have;
![m^{-1}(x) = \dfrac{17}{2} \pm \sqrt{x + \dfrac{289}{4} }}](https://tex.z-dn.net/?f=m%5E%7B-1%7D%28x%29%20%3D%20%5Cdfrac%7B17%7D%7B2%7D%20%5Cpm%20%5Csqrt%7Bx%20%2B%20%5Cdfrac%7B289%7D%7B4%7D%20%7D%7D)
We note that the maximum or minimum point of the function, m(x) = x² - 17·x found by differentiating the function and equating the result to zero, gives;
m'(x) = 2·x - 17 = 0
x = 17/2
Similarly, the second derivative is taken to determine if the given point is a maximum or minimum point as follows;
m''(x) = 2 > 0, therefore, the point is a minimum point on the graph
Therefore, as x increases past the minimum point of 17/2, m⁻¹(x) increases to give;
to increase m⁻¹(x) above the minimum.