The graph of the continuous function g, the derivative of the function f, is shown above. the function g is piecewise linear for -5 ≤ x < 3 and g(x) = 2(x-4)^2 for 3 ≤ x ≤ 6. (a) if f(1) = 3, what is the value of f(-5) ? (b) evaluate . (c) for -5 < x < 6, on what open intervals, if any, is the graph of f both increasing and concave up? give a reason for your answer. (d) find the x-coordinate of each point of inflection of the graph of f. give a reason for your answer. (this was on the no-calculator section of the recently-released ap calculus ab 2018 exam so i appreciate it if you tried to limit calculator usage)

A) is continuous, so is also continuous. This means if we were to integrate , the same constant of integration would apply across its entire domain. Over , we have . This means that



For to be continuous, we need the limit as to match . This means we must have



Now, over , we have , so , which means .

b) Integrating over [1, 3] is easy; it's just the area of a 2x2 square. So,



c) is increasing when , and concave upward when , i.e. when is also increasing.

We have over the intervals and . We can additionally see that only on and .

d) Inflection points occur when , and at such a point, to either side the sign of the second derivative changes. We see this happening at , for which , and to the left of we have decreasing, then increasing along the other side.

We also have along the interval , but even if we were to allow an entire interval as a "site of inflection", we can see that to either side, so concavity would not change.