Article in English: This is just like our example in the first section, except that we're dealing with a cubic equation rather than a quadratic equation, so we can solve it in the same way.  First, we'll take our equation's derivative:s = 5t3 - 3t2 + 2t + 9s = (3)5t(3 - 1) - (2)3t(2 - 1) + (1)2t(1 - 1) + (0)9t0 - 115t(2) - 6t(1) + 2t(0)15t(2) - 6t + 2  Then, we'll plug in our value for t (4):s = 15t(2) - 6t + 215(4)(2) - 6(4) + 215(16) - 6(4) + 2240 - 24 + 2 = 218 meters/second For this problem, we'll use (1,3) as our P point, but we'll have to find a few other points near it to use as our Q points. Then, it's just a matter of finding our H values and making an estimation.  First, let's find Q points at t = 2, 1.5, 1.1 and 1.01.s = 4t2 - tt = 2: s = 4(2)2 - (2)4(4) - 2 = 16 - 2 = 14, so Q = (2,14)t = 1.5: s = 4(1.5)2 - (1.5)4(2.25) - 1.5 = 9 - 1.5 = 7.5, so Q = (1.5,7.5)t = 1.1: s = 4(1.1)2 - (1.1)4(1.21) - 1.1 = 4.84 - 1.1 = 3.74, so Q = (1.1,3.74)t = 1.01: s = 4(1.01)2 - (1.01)4(1.0201) - 1.01 = 4.0804 - 1.01 = 3.0704, so Q = (1.01,3.0704)  Next, let's get our H values:Q = (2,14): H = (14 - 3)/(2 - 1)H = (11)/(1) = 11Q = (1.5,7.5): H = (7.5 - 3)/(1.5 - 1)H = (4.5)/(.5) = 9Q = (1.1,3.74): H = (3.74 - 3)/(1.1 - 1)H = (.74)/(.1) = 7.3Q = (1.01,3.0704): H = (3.0704 - 3)/(1.01 - 1)H = (.0704)/(.01) = 7.04  Since our H values seem to be getting very close to 7, we can say that 7 meters/second is a good estimate for the instantaneous velocity at (1,3).

Summary in Vietnamese:
Tìm vận tốc tức thời khi t = 1 với phương trình quãng đường dịch chuyển là s = 5t3 - 3t2 + 2t + 9. Sử dụng phương pháp ước lượng bằng đồ thị để tìm vận tốc tức thời tại (1;3) cho phương trình quãng đường dịch chuyển s = 4t2 - t.