I have to carry out a few steps to show that 2 < e < 3. (e as in 2.71828..)(adsbygoogle = window.adsbygoogle || []).push({});

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?

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# 2 < e < 3

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