If f(1) = 9 and f '(x) ≥ 3 for 1 ≤ x ≤ 6, how small can f(6) possibly be?

Question

If f(1) = 9 and f '(x) ≥ 3 for 1 ≤ x ≤ 6, how small can f(6) possibly be?

Answer 1

Another way of thinking about this that doesn't use integration is to think of the initial point at (1, 9) and that fact that the derivative function is at smallest equal to 3, meaning the slope is at smallest equal to 3. To go over 5 units to 6 means you would need to go up at least 3(5) or 15 units which is 15 + 9 or 24. So the smallest value for x = 6 is 24.

Answer 2

24 is the minimum possible for f(6)

f''(x) > 3

integrate to get f(x) > 3x + c

f(1) = 9= 3x + c

c=6

f(x) > 3x + 6

f(6) > 3(6) + 6 = 24

Answer 3

f(x) = ∫f'(x)dx ≥ 3x + C; f(1) ≥ 3 + C ≤ 9 ; C ≤ 6;

f(x) ≥ 3x + 6; f(6) ≥ 3·6 + 6; f(6) ≤ 24

Answer 4

The smallest f'(x) can be from 1 to 6 is 3. So the minimum value at x = 6 is found by inputting 6 into the line equation with slope 3 that goes through the point (1,9). f(x) = f(1) + 3(6-1)