Answer:
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
