If f(3) = 11 and f '(x) ≥ 2 for 3 ≤ x ≤ 8, how small can f(8) possibly be?

So f'(x) is basically a slope. Since we want f(8) to be as small as possible, we would want smallest slope from x=3 to x=8, right?

f'(x) can only be 2 and not less so then f(x) would increase by 2 for every x increasing by 1 all the way to x=8.

So then smallest possible f(8) would be f(8) = f(3) + 2 + 2 + 2 + 2 + 2 = f(3) + 2(8-3) = 21

Hope this helps.

Here are the answers to the ones that i could see.