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)
a = 1, b = -12, c = 58 - x
Therefore;
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).
