Problem of the Week #14

Problem 14: 1998-1999

Given , where f(t) is the graph shown above.
(a)    Evaluate g(-3) and g(3).
(b)    Estimate g(-2), g(-1), and g(0).
(c)    On what interval is g increasing?
(d)    Where does g have a maximum value?
(e)    Sketch a rough graph of g(x).
(f)    Use the graph of g(x) in part (e) to sketch the graph of g'(x).

Difficulty rating:

Solution: (a) To evaluate g(-3), we plug -3 into the function for x, noting that g is a function of x, to get:

which clearly equals 0, since the upper and lower integration limits are equal. In order to find g(3), we note that the function is origin-symmetric and therefore odd. Thus, by a property of integration, since the integration limits are opposites and the function is odd, the value of g(3) is 0 also.

(b) In order to approximate these values, we will need to approximate the area beneath f(t), as g gets its value from accumulating area beneath g, thus making it an accumulation function. The easiest way to do this is simply to count boxes beneath the graph! Each box represents one square unit of area.

Note that in the above graph, the shaded area represents g(-2), the area beneath f(t) between -3 and -2, as defined by the function. The shaded area is approximately 1 full box, if not a teensy bit more. Since we are approximating, I will say g(-2) = 1. In the same fashion, g(-1) = 3.5 and g(0) = 5.5 (both approximate).

(c) The value of g will increase whenever I continue to gather positive signed area beneath f(t), which will happen until I pass x = 0. Once I start plugging in values of x > 0 into g, I will begin to accumulate negative signed area, or area beneath the x-axis. This area will eat away at my value of g until I reach x = 3, when all the area I have gathered is completely negated by the negative area (because the function is odd, as noted in part (a)) and g(3) = 0 as noted above. Therefore, g is increasing on (-3,0).

(d) If g changes from increasing to decreasing at x = 0, then g has a max at x = 0. If you want the actual point on g where the max occurs, we found it in part (b): (0, 5.5).

(e) and (f) are submitted by celebrity guest Brenna Hogan via email:

Note that g'(x) is equal to f(x), as guaranteed by the Second Fundamental Theorem of Calculus.

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