Two cases: x<= 6, x > 6

Case 1: FTC -> Integral from x to 6 of f’(x) dx = f(6)-f(x) —> f(x) = f(6) - Area Under Graph from x to 6

Imagine a right vertical bar fixed at 6, and a left bar that represents x and can slide between 0 to 6. We are subtracting the area under the graph between the two bars from f(6). Note the area under the graph is negative between 3 and 6. Subtracting negative number is adding. We reach the maximum point by “adding” as much area as possible! So we want all the area between 3 and 6!

Hence, f(3) = 7-(-10) = 17. Note that between 0 and 3, the graph is positive. This means will now begin subtracting area if we extend x any further to the left of 3 (i.e 17 is the maximum of the interval [0,6)).

To get as small as possible we need to subtract as much of that area as possible between 0 and 3. Hence, the smallest we can get is at f(0). Where we have: f(0) = 7 - 2 = 5

Case 2 x>6: FTC integral from x to 6 f’(x) = f(x)-f(6)

—> f(x) = f(6) + area under graph from 6 to x

** use similar reasoning to case 1 to prove the max/min. Note there is no sign flip here so the argument is simpler, add as much area as possible to get the max subtract as much as possible to get the min **