Question

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

Which represents a domain restriction and the corresponding inverse function?
O x26; h'(x) = 6 +Vx-22
O x26; h-'(x) = 6 - Vx-22
O x2 12; h-1(x) = 12 +/x-58
0 x : 12: h-1(x) = 12 - x -58

Answer 1

the answer is A

Explanation:

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Answer 2

The correct option is;

x ≥ 6; h⁻¹(x) = 6 +√(x - 22)

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 = \frac{-b\pm \sqrt{b^(2)-4\cdot a\cdot c}}{2\cdot a}`

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

Therefore;

`x = \frac{-(-12)\pm \sqrt{(-12)^(2)-4* (1)* (58 - x)}}{2* (1)}= (12\pm √(144-232 + 4 * x)))/(2)`

`x = (12\pm √(4 * x - 88))/(2) = (12\pm √(4 * (x - 22)))/(2) = (12\pm 2 * √( (x - 22)))/(2)`

`x = (12\pm 2 * √( (x - 22)))/(2) = {6 \pm √( (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).