Question

If f(1) = 15 and f '(x) ≥ 1 for 1 ≤ x ≤ 5, how small can f(5) possibly be

Answer 1

19

Explanation:

f'(c) = (f(b) - f(a))/(b - a)

In this case, we have a = 1 and b = 5, so we can write:

f'(c) = (f(5) - f(1))/(5 - 1)

Solving for f(5), we get:

f(5) = f(1) + f'(c)(5 - 1)

Since we know that f '(x) ≥ 1 for 1 ≤ x ≤ 5, we have:

f(5) = f(1) + f'(c)(5 - 1) ≥ 15 + 1(5 - 1) = 19

Therefore, the smallest value that f(5) can possibly be is 19.

*IG:whis.sama_ent