If f(2) = 12 and f '(x) ≥ 3 for 2 ≤ x ≤ 5, how small can f(5) possibly be?

21

Step-by-step explanation:

From our previous knowledge, if f'(x) = m ,

Then f(x) = mx+c

For a smaller value of f(5), m should be the smallest possible integer if m = 3

i.e. f(x) = 3x+c

replacing the value of x with 2 given from the question to get the value of c, we have:

f(x) = 3x+c

f(2) = (3×2)+c

where; f(2) = 12

12 = 6+c

- c = -12 + 6

- c = - 6

c = 6

Now, f(x) can now be;

f(x) = 3x + 6

f(5) = (3×5)+6

f(5) = 15 + 6

f(5) = 21

Therefore, smallest possible value of f(5) is 21

$27.5

step-by-step explanation:

$110.00 divided by 4