AB1: Let g be the function defined by g(x) = {f(x^2 - 5), x <= 3; 5 - 2x, x > 3. The function f is twice differentiable on the closed interval [-2,10] and satisfies f(6) = -4. The graph of f', the derivative of f, is shown in the figure above. The graph of f' has horizontal tangent lines at x = 2, x = 5, x = 8, and x = 9. The areas of the regions between the graph of f' and the x axis are labeled in the figure.
(a) Show that g is continuous at x = 3.
(b) Find the absolute maximum value of f on the interval [-2,10]. Justify your answer.
(c) For x != 3, the function k is defined by k(x) = (3 integral from 4 to 3x f'(t)dt - 4x) / (3e^(2f(x)+5) - x). It is known that lim_{x->3} k(x) can be evaluated using L'Hospital's Rule. Find f(3) and evaluate lim_{x->3} k(x).
(d) Let y = h(x) be a function such that dy/dx = f'(2x) * (3y - 1) where y > 1/3. Find d^2y/dx^2 and use this expression to determine if the graph of h(x) is concave up or down at h(1) = 2.
(e) Consider the function y = h(x) from part (d). For -1 < x < 5, on what interval(s) is the function h(x) increasing? Give a reason for your answer.
(f) Consider the differential equation from part (d). Use separation of variables to find a general solution for y = h(x) of the form ln|3y - 1| = af(bx) + C.