a. To find the x-values such that the tangent line to the graph of f at (x, f(x)) is horizontal, we need to find where the derivative of f is equal to zero. Given that f'(x) is the first derivative of f, we can use the information given in property (ii) to determine that f'(2) = 0.
So, one x-value where the tangent line is horizontal is x = 2.
b. To find an expression for f(x), we need to integrate the second derivative of f. Given that f''(x) = 24x - 18, we can integrate it to get the first derivative, f'(x):
∫ f''(x) dx = ∫ (24x - 18) dx
f'(x) = 12x^2 - 18x + C1
Next, we can use the information given in property (i) to determine f'(1) = -6. Plugging this into the expression for f'(x), we get:
-6 = 12(1)^2 - 18(1) + C1
-6 = 12 - 18 + C1
C1 = -6 - 12 + 18
C1 = 0
Now, integrating f'(x), we get f(x):
∫ f'(x) dx = ∫ (12x^2 - 18x + 0) dx
f(x) = 4x^3 - 9x^2 + C2
To find the expression for f(x), we need to determine the value of C2. Using the information in property (iii), we have f'(2) = 0:
0 = 4(2)^3 - 9(2)^2 + C2
0 = 32 - 36 + C2
C2 = 36 - 32
C2 = 4
So, the expression for f(x) is:
f(x) = 4x^3 - 9x^2 + 4