If f(3) = 5 and f′(t)≥3 for 3≤t≤5, how small can f(5) possibly be

Answer 1

f'(t) >= 3 for 3 <= t <= 5 means

there is at least 3Δt increase in f(t) for each Δt increase in t from t = 3 to t = 5

So for f(5)_min, f'(t) should be min too, which is 3,

f(5)_min = f(3 + 2)_min = f(3) + 3*2 = 5 + 6 = 11

Answer 2

Since f'(t) ≥ 3 on [3,5], f'(t) > 0 on [3,5].

So, f(t) is increasing on the interval [3,5].

But, f(3) = 5. Therefore, f(5) must be greater than 5.