The graph shows the derivative of a function 𝑓. Given that 𝑓 of three is equal to two, what is 𝑓 of four?
We need to be really careful here. We’ve not been given the graph of the function 𝑓 but the graph of its derivative. That’s the graph of 𝑓 prime. And so, we can’t simply read values straight off the graph. Instead, we need to find a way to link 𝑓 prime with its values for 𝑓. We’re going to use the second part of the fundamental theorem of calculus. And this says that if 𝑓 is continuous on the closed interval 𝑎 to 𝑏, then the definite integral between 𝑎 and 𝑏 of 𝑓 of 𝑥 with respect to 𝑥 is equal to capital 𝐹 of 𝑏 minus capital 𝐹 of 𝑎, where capital 𝐹 is the antiderivative of 𝑓. That’s a function such that capital 𝐹 prime is equal to 𝑓. Now, actually, we can reword this a little. Since 𝑓 is the antiderivative of 𝑓 prime, we can say that the definite integral between 𝑎 and 𝑏 of 𝑓 prime of 𝑥 with respect to 𝑥 is equal to 𝑓 of 𝑏 minus 𝑓 of 𝑎.
Now, before we apply this theorem, let’s just check the continuity of our function 𝑓. Well, we’re given a graph of its derivative and we know that any differentiable function must be continuous at every point in its domain. So, simply because we’ve been given the derivative of 𝑓, we know 𝑓 itself must be continuous. Now, we’re given 𝑓 of three, and we’re looking to find 𝑓 of four. So, we’re going to define 𝑎 as being equal to three and 𝑏 being equal to four. And this means then that the definite integral between three and four of 𝑓 prime of 𝑥 with respect to 𝑥 must be equal to 𝑓 of four minus 𝑓 of three. Well, 𝑓 of four is what we’re trying to find. But we’re told in the question that 𝑓 of three is equal to two.
And what about the definite integral between three and four of 𝑓 prime of 𝑥 with respect to 𝑥? Well, to work this out, we need to go to the graph. We know that the definite integral between two points of a function represents the area under a curve. And if that area lies below the 𝑥-axis, that will come out as a negative value. We’re looking to find the area between the curve and the 𝑥-axis between the points 𝑥 equals three and 𝑥 equals four. Well, by dropping the vertical lines 𝑥 equals three and 𝑥 equals four onto our diagram, we see that we need to find the area of a trapezium. So, we can recall the area of a trapezium as being a half times 𝑎 plus 𝑏 times ℎ, where 𝑎 and 𝑏 are the lengths of the parallel sides and ℎ is the height between them.
Well, our two parallel sides are two units in length and four units in length. That’s essentially the absolute value of the value of the derivative of those points. The distance between them is one, and we know that the integral will give us a negative result since this trapezium lies below the 𝑥-axis. Well, two plus four is six, and six times one is six. So, we have negative one-half times six equals 𝑓 of four minus two or negative three equals 𝑓 of four minus two. Remember, we’re trying to find the value of 𝑓 of four. So, we’re going to solve for 𝑓 of four by adding two to both sides. 𝑓 of four is therefore equal to negative three plus two, which is negative one. And so, by applying the second part of the fundamental theorem of calculus to the information we’ve been given, we found 𝑓 of four to be equal to negative one.