Practice Answers

Practice 1: g is increasing on [2,4] and [6,8].
g is decreasing on [0,2] and [4,5],
g is constant on [5,6].


Practice 2: The graph in Fig. 33 shows the rate of population change, dR/dt.

Fig. 33


Practice 3: The graph of f is shown in Fig. 34. Notice how the graph of f is 0 where f has a maximum and minimum.

Fig. 34


Practice 4: The Second Shape Theorem for helicopters:
(i) If the upward velocity h is positive during time interval I then the height h is increasing during time interval I.
(ii) If the upward velocity h is negative during time interval I then the height h is decreasing during time interval I.
(iii) If the upward velocity h is zero during time interval I then the height h is constant during time interval I.


Practice 5: A graph satisfying the conditions in the table is shown in Fig. 35.

x210123f(x)112102f(x)101211

Fig. 35


Practice 6: f(x)=x33x224x+5.

f(x)=3x26x24=3(x4)(x+2).

f(x)=0 if x=2,4.

If x<2
then f(x)=3(x4)(x+2)=3 (negative) (negative) >0 so f is increasing.

If 2<x<4
then f(x)=3(x4)(x+2)=3 (negative)(positive) <0 so f is decreasing.

If x>4
then f(x)=3(x4)(x+2)=3 (positive) (positive) >0 so f is increasing

f has a relative maximum at x=2 and a relative minimum at x=4.
The graph of f is shown in Fig. 36.

Fig. 36


Practice 7: Fig. 37 shows several possible graphs for g. Each of the g graphs has the correct shape to give the graph of g. Notice that the graphs of g are "parallel," differ by a constant.

Fig. 37


Practice 8: Fig. 38 shows the height graph for the balloon. The balloon was highest at 4pm and had a local minimum at 6pm.

Fig. 38


Practice 9:
f(x)=3x212x+7 so f(x)=6x12.

f(x)=0 if x=2.
If x<2, then f(x)<0 and f is decreasing.
If x>2, then f(x)>0 and f is increasing.
From this we can conclude that f has a minimum when x=2 and has a shape similar to Fig. 19(b).

We could also notice that the graph of the quadratic f(x)=3x212x+7 is an upward opening parabola. The graph of f is shown in Fig. 39.

Fig. 39