I have to carry out a few steps to show that 2 < e < 3. (e as in 2.71828..) Let f(t) = 1/t for t > 0. (a) Show that the area under y = f(t), above y = 0, and between t = 1 and t = 2 is less than 1 square unit. Deduce that e < 2. This is easy, just integrating and getting ln(2). From graph I see that ln(2) < 1 => e > 2. (b) Show that all tangent lines to the graph of f lie below the graph. Also not very hard; just showing that f'(t) < 0 gives that it's a decreasing function in the interval ]0,inf[, so the tangent lines have negative slopes, hence lies below the graph. (c) Find the lines T_2 and T_3 that are tangent to y = f(t) at t = 2 and t = 3, respectively. I got T_2: y = (-1/4)t + 1 T_3: y = (-1/9)t + (2/3) (d) Find the area A_2 under T_2, above y = 0, and between t = 1 and t = 2. Also find the area A_3 under T_3, above y = 0, and between t = 2 and t = 3. I got A_2 = 5/8 A_3 = 7/18 (e) Show that A_2 + A_3 > 1 square unit. Deduce that e < 3. OK, so A_2 + A_3 = (5/8) + (7/18) = 73/72 > 1, but I can't see how this leads to e < 3. Any tips or suggestions?