Science:Math Exam Resources/Courses/MATH100/December 2011/Question 01 (l)/Solution 1

We are being told that the instantaneous rate of change of the function is at least of 3 on the interval [-1,2], so the smallest possible value of ƒ(2) is the case where the function has the smallest rate of change, that is a constant rate of change of 3 which means it is a straight line (of slope 3).

And since we are starting at the point (-1,9) with a slope of 3 and a run of 3 (from -1 to 2) we will end up with a rise of 9 and so the minimal value of ƒ(2) is 18.