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Solutions to the exercises

Exercise 1
a

6

b

f ' ( x ) = 2 x + 4

c

f ' ( 1 ) = 2 1 + 4 = 6

d

At that point, the graph of f has a horizontal tangent. This means that the graph of f has a minimum at that point.

e

f ' ( 0 ) = 4 and f ' ( - 4 ) = - 4

f

f ' ( x ) = 2 x + 4 = 2 gives you x = - 1 , so at ( - 1 , - 3 ) .

Exercise 2
a

49 m/s

b

Δ s Δ t = 4,9 ( 10 + h ) 2 - 4,9 10 2 h and therefore s ' ( 10 ) = 98 m/s.

c

v ( t ) = s ' ( t ) = 9,8 t

d

120 km/h = 33 1 3 m/s and s ' ( t ) = 9,8 t = 33 1 3 gives you t 3,40 s.

Exercise 3

Δ y Δ x = c - c h = 0 for every h 0

Exercise 4
a

T P ' ( q ) = 900 - 120 q

b

T P ' ( 5 ) is the speed with which the profit increases (or decreases) as q increases.

c

T P ' ( q ) = 900 - 120 q = 0 as q = 7,5 .
The graph of T P is an open down parabola and therefore has a maximum at q = 7,5 .
The maximum profit is therefore seen at sales of 750 cars per year.

Exercise 5
a

Δ H Δ t = H ( 10 ) - H ( 0 ) 10 -1,79 mg/L per day.

b

H ' ( 0 ) -4,46 and H ' ( 4 ) -1,82 mg/L per day.
The speed of degradation is getting smaller because the graph is decreasing less rapidly.

c

You cannot rearrange Δ H Δ t = 20 0,8 ( t + h ) - 20 0,8 t h such that it would be possible to divide by h . This means that you cannot calculate the limit for h 0 at this point.

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