2014 AB FRQ #3 By Philip Schnee

For this problem, you are given that g(x) is the integral of f(x), which is graphed, from -3 to x. To find g(3), you need to integrate the graph above from 0 to 3. You are integrating, so any area below the x axis is negative. You are still finding the area under the line from -3 to 3, but keep this in mind. To find this area, you need to visualize 3 right triangles on the graph, find their areas with A=(½)B x H, and add them to get your final answer. Graph of f (½) x 3 x 4 = 6 ( ½) x 2 x 4 = 4 (½) x 1 x 2 = 1, but it’s negative! = 9 The correct answer is g(3) = 9.

For part (b), you again need to remember that they gave you the fact that g’(x) = f(x). So, to find when g is both increasing and concave down, you need to find the intervals in which f(x) is positive and f’(x) is negative. Remember, derivative is slope! Once you realize this, the rest is easy and quick. Simply look and the graph, which is already on the interval -5 < x < 4, like the question is asking. f(x) is positive from -5 to 2, because it’s above the x-axis. f’(x), the slope of this graph, is negative from -5 to -3, and from 0 to 2. So, now you know that f(x) is positive and f’(x) is negative on the intervals -5 < x < 3, and 0 < x < 2. That’s the correct answer. Quick and painless, right?

Are you thinking what I’m thinking? That’s right, quotient rule! Remember, it’s (u’v-uv’) / v 2. So look at the problem, us are on top and vs on bottom. What you need has magically appeared above the question. Let’s get started by finding an equation for h’(x). On the top, you will have: (g’(x)) (5x) - (g(x)) (5). On the bottom, you will have (5x) 2, which equals 25x 2. Now you just have to fill in the numbers. Remember, x is 3 in this situation. From part (a), you already know that g(3) = 9. So, the only thing left to find is g’(3), but remember, g’(x) = f(x)! Look on the original graph for what f(3) is. As you can see, it’s -2. Now all that’s left in the problem is simple math. U’ = g’(x)V’ = 5U = g(x)V= 5x f(3) = -2

So, you have [ (g’(x)) (5x) - (g(x)) (5) ] / 25x 2. Now, just fill everything in, remember that x is 3. By doing that, you get [ (-2)(5)(3) - (5)(9) ] / (25)(9) Do your simple multiplication, and you end up getting -75 / 225. Simplify, and you get your final answer as h’(3) = -⅓. Part (c)

Do rise over run. (-4 - 0) / (4 - 2) = -2 f’(2) = -2. This problem won’t be as bad as the last one, I promise. If you still remember that derivative is slope, you need to find p’(x) to get the slope of the line tangent to p(x). With the given equation, p’(x) = [ f’(x 2 -x) ] (2x-1). The 2x-1 is from the chain rule, since it’s the derivative of x 2 -x. You know that x=-1, so just fill in -1 for all the x’s in the equation. If you do that, you should get [ f’(1--1) ] (-2-1), which gets you to [ f’(2) ] (-3). From the given graph, you can find f’(2) by doing old- school slope. Now that you know f’(2) = -2, you can do -2 x -3 = -6. That’s the slope of the tangent line, which is all they are asking. So, your answer is p’(- 1) = 6, and now you’re all done!