Question

Consider the function h(x) =x2 – 12x + 58.

Answer 1

Answer:

The correct option is;

x ≥ 6; h⁻¹(x) = 6 +√(x - 22)

Step-by-step explanation:

Given the function, h(x) = x² - 12·x + 58

We can write, for simplification;

y = x² - 12·x + 58

Therefore;

When we put x as y to find the inverse in terms of x, we get;

x = y² - 12·y + 58

Which gives;

x - x = y² - 12·y + 58 - x

0 = y² - 12·y + (58 - x)

Solving the above equation with the quadratic formula, we get;

0 = y² - 12·y + (58 - x)

$x = \dfrac{-b\pm \sqrt{b^{2}-4\cdot a\cdot c}}{2\cdot a}$

a = 1, b = -12, c = 58 - x

Therefore;

$x = \dfrac{-(-12)\pm \sqrt{(-12)^{2}-4\times (1)\times (58 - x)}}{2\times (1)}= \dfrac{12\pm \sqrt{144-232 + 4 \times x)}}{2}$

$x = \dfrac{12\pm \sqrt{4 \times x - 88}}{2} = \dfrac{12\pm \sqrt{4 \times (x - 22)}}{2} = \dfrac{12\pm 2 \times \sqrt{ (x - 22)}}{2}$

$x = \dfrac{12\pm 2 \times \sqrt{ (x - 22)}}{2} = {6 \pm \sqrt{ (x - 22)}}$

We note that for the function, h(x) = x² - 12·x + 58, has no real roots and the real minimum value of y is at x = 6, where y = 22 by differentiation as follows;

At minimum, h'(x) = 0 = 2·x - 12

x = 12/2 = 6

Therefore;

h(6) = 6² - 12×6 + 58 = 22

Which gives the coordinate of the minimum point as (6, 22) whereby the minimum value of y = 22 which gives √(x - 22) is always increasing

Therefore, for x ≥ 6, y or h⁻¹(x) = 6 +√(x - 22)  and not 6 -√(x - 22) because 6 -√(x - 22) is less than 6

The correct option is  x ≥ 6; h⁻¹(x) = 6 +√(x - 22).