2011 Calculus AB free response #4c

Part C. Find all values of x on the interval negative 4 is less than x is less than 3 for which the graph of g has a point of inflection. Give a reason for your answer. So an inflection point is a point where the sign of the second derivative changes. So if you take the second derivative at that point, or as we go close to that point, or as we cross that point, it goes from positive to negative or negative to positive. And to think about that visually, you could think of some examples. So if you have a curve that looks something like this, you'll notice that over here the slope is negative, but it's increasing. It's getting less negative, less negative. Then it goes to 0. Then it keeps increasing. Slope is increasing, increasing, all the way to there, and then it starts getting less positive. So it starts decreasing. So it's increasing. The slope is increasing over at this point right over here. So even though the slope is negative, it's getting less negative over here. So it's increasing. And then the slope keeps increasing. It keeps on getting more and more positive up to about this point. And then the slope is positive, but then it becomes less positive. So the slope begins decreasing after that. So this right over here is a point of inflection. The slope has gone from increasing to decreasing. And if the other thing happened, if the slope went from decreasing to increasing, that would also be a point of inflection. So if this was maybe some type of a trigonometric curve, then you might see something like this. And so this also would it be a point of inflection. But for this, our g of x is kind of hard to visualize the way they've defined it right over here. So the best way to think about it is just figure out where its second derivative has a sign change. And to think about that, we have to find its second derivative. So let's write g of x over here. We know g of x is equal to 2x plus the definite integral from 0 to x of f of t dt. We've already taken its derivative, but we'll do it again. g prime of x is equal to 2 plus-- fundamental theorem of calculus. The derivative of this right over here is just f of x. And if we have the second derivative of g-- g prime prime of x-- this is equal to-- derivative of 2 is just 0. And then derivative of f of x is f prime of x. So asking where this has a sign change, asking where our second derivative has a sign change, is equivalent to asking where does the first derivative of f have a sign change? And asking where the first derivative of f has a sign change is equivalent to saying where does the slope of f have a sign change? You can view this as the slope or the instantaneous slope of f. So we want to know when the slope of f has a sign change. So let's think about. Over here, the slope is positive. It's going up. It's up. It's increasing, but it's positive. And that's what we care about. So let's write it. I'll do it in green. So the slope is positive this entire time. It's increasing. It's increasing. It's positive. It's getting less positive now. It's starting to decrease, but the slope is still positive. The slope is still positive all the way until we get right over there. You can see it gets pretty close to zero. And then the slope gets a negative. And then right over here, the slope is negative. The slope is negative right over here. So this is interesting. Because even though f is actually not differentiable right here-- so f is not differentiable at that point right over there. And you could see it because the slope goes pretty close to 0, and then it just jumps to negative 3. So you have a discontinuity of the derivative right over there, but we do have a sign change. We go from having a positive slope on this part of the curve to having a negative slope over this part of the curve. So we experience a sign change right over here at x is equal to 0, a sign change in the first derivative of f, which is the same thing as saying a sign change in the second derivative of g. And a sign change in the second derivative of g tells us that, when x is equal to 0, the graph of g has a point of inflection.