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12345In graphs

Solutions to the exercises

Exercise 1

, - 1 : decelerating increase
- 1 , 0 : accelerating decrease
0 , 1 : decelerating decrease
1 , : accelerating increase
max. f ( - 1 ) 4 and min. f ( 1 ) - 4
At point ( 0 , 0 ) the decrease is fastest, since the slope of the curve is steepest at that point.

Exercise 2

, - 3 : decelerating decrease
- 3 , 0 : accelerating increase
0 , 3 : decelerating increase
3 , : accelerating decrease
max. f ( 3 ) = 27 and min. f ( - 3 ) = - 27
At point ( 0 , 0 ) the increase is fastest since the slope of the curve is steepest at that point.

Exercise 3
a

max. f ( 0 ) = 8 , min. f ( - 2 ) = 0 and min. f ( 2 ) = 0

b

0 , 1

c

B f = 0 ,

Exercise 4
a

He opened his parachute after 60 seconds, because at that point there is a kink in the curve followed by a constant decrease.

b

The speed of falling is increasing.

c

The graph is a straight line during that interval. v = 1000 / 10 = 10 m/s.

Exercise 5

The function has a maximum at 2.30 p.m., since the curve moves from increasing to decreasing at that point.

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