A function is defined by f(x)=x/(2x-2)^(1/2): (a)Determine the maximum domain of f. (b)Differentiate f. (c)Find the inflection points of the function's graph.

This is a common A-level question tackling functions and differentiation. (a) First, it should be noted that the function f is defined within the field of real numbers and thus the value of 2x-2 in the square root must be positive which leads to a lower bound, x>1. Therefore, the domain of f is {x ∈ ℝ : x >1}.(b) Differentiation should be done in our case according to the quotient rule (if f=g/h, g and h are differentiable functions and h(x) not 0, then f'=(g'h-gh')/h²): f'(x)=((2x-2)^1/2-0.5x(2x-2)^-1/2*2)/(2x-2)f'(x)=(2x-2-x)/(2x-2)^3/2.(c) The inflection point is the point where a function changes its curve from concave to convex or vice versa. It is the solution to the equation f"(x)=0. The second derivative of f, f"(x)=((2x-2)^3/2-3/2(x-2)(2x-2)^1/2*2)/(2x-2)³ and since the denominator cannot be 0 the equation becomes:(2x-2)^1/2((2x-2)-3(x-2))=0(2x-2)^1/2(4-x)=0The solutions are 1 and 4 but at (a) we determined that x>1 so x=4 is the only inflection point of the function.