The graph of a function f is shown below:

(a)    Evaluate .

(b)    Determine the average value of the function on the interval [1,7].

(c)    Determine the answers to parts (a) and (b) above for the function f(x)+2.

(d)    Determine where on [1,7] the instantaneous rate of change of f equals the average rate of change.

Difficulty rating: 

(a)     The integral  represents the signed area bounded by the graph of f and the x-axis (the presence of a definite integral and a single function always implies the x-axis is a boundary). So, break the area beneath the graph of f into smaller geometric shapes whose area is easy to find (i.e. triangles and rectangles) as shown in the figure below.

(b)    The average value is given by a formula you must make sure to memorize:

(c)    Adding two to the function has the effect of moving the graph up two places, resulting in the lower rectangle as pictured in part (a) changing in width from 1 to 3, thus making =20; the rectangle was increased in area from 6 to 18 square units. Using the formula in part (b), we get that the average value is now 10/3.

(d)    The average rate of change of the function is the slope of the secant line connecting the two endpoints of the interval in question. In this problem, the points are (1,3) and (7,1). The slope of the line joining these two points (change in y/change in x) is -1/3. Now, the instantaneous rate of change is equal to the slope not of a secant line, but of a tangent line. Because the graph of f in this problem is linear (although it's in pieces), the slope of the tangent line is the slope of the line itself (what other line could be tangent to a line other than the line itself?). By observation, the slopes of the line segments in f are (left to right): -2, 1, -1, and 0.

        Wait a minute ... I thought the Mean Value Theorem said that somewhere on a closed interval the average rate of change equals the instantaneous rate of change! That's not true in this case ... -2, 1, -1, and 0 are certainly not equal to -1/3. Why doesn't it work? Remember that in order for the Mean Value Theorem to be applied, the function must first be continuous (which f is) and differentiable throughout the whole interval (which f is not since there are sharp points on f at x = 2, 3, and 4, negating the possible presence of a derivative). The Mean Value Theorem cannot be applied to this problem, so it's a trick question!

        Don't forget: a derivative does not exist on a function at a point if (1) the function is discontinuous there, (2) the function has a sharp point there, and (3) if the tangent line to the graph is vertical there. Since a vertical line has no slope, the function cannot have a derivative there (since the derivative is a slope).

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