Read this section to learn how the first derivative is used to determine the shape of functions. Work through practice problems 1-9.
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.
x−2−10123f(x)1−1−2−102f′(x)−1012−11
Fig. 35
Practice 6: f(x)=x3−3x2−24x+5.
f′(x)=3x2−6x−24=3(x−4)(x+2).
f′(x)=0 if x=−2,4.
If x<−2
then f′(x)=3(x−4)(x+2)=3 (negative) (negative) >0 so f is increasing.
If −2<x<4
then f′(x)=3(x−4)(x+2)=3 (negative)(positive) <0 so f is decreasing.
If x>4
then f′(x)=3(x−4)(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)=3x2−12x+7 so f′(x)=6x−12.
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)=3x2−12x+7 is an upward opening parabola. The graph of f is shown in Fig. 39.
Fig. 39