(a) The particle is moving to the right when 0 &lt

5. Let f (x) be the given function. In page 147, we know that the slope of the tangent line at (3, 2) is exactly equal to

f⁰(3) = lim

x→3

f (x) − f (3) x − 3 Notice that

f (x) − f (3)

x − 3 = 1

x − 3

 x − 1 x − 2 − 2



= 1

x − 3

 3 − x x − 2



= −1 x − 2 so that

f⁰(3) = lim

x→3

f (x) − f (3)

x − 3 = lim

x→3

−1

x − 2 = −1 The tangent line of y = f (x) at (3, 2) is

y − 2 = (−1)(x − 3)

11. (a) The particle is moving to the right when 0 < t < 1 and 4 < t < 6, to the left when 2 < t < 3, and standing when 1 < t < 2, 3 < t < 4.

(b) See page A72.

17. Recall that given a function y = g(x) in the plane. g⁰(a) means the slope of the tangent line at (a, g(a)) if g⁰(a) exists. The answer is

g⁰(0) < 0 < g⁰(4) < g⁰(2) < g⁰(−2)

19. We only to draw a function with f (0) = 0, and the slope of its tangentline at x = 0 is 3, at x = 1 is 0 and −1 as x = 2. The following gives an example of such function.

Figure 1: y = x³/3 − 2x² + 3x

43. (a) By definition,

avg₁ = C(105) − C(100)

105 − 100 = 26405/4 − 6500

105 − 100 = 20.25 The other is similar.

avg₂ = C(101) − C(100)

101 − 100 = 20.05 1

(b) The rate from x to 100 is given by

C(x) − C(100) x − 100 the instantaneous rate is given by taking limit.

x→100lim

C(x) − C(100)

x − 100 = lim

x→100

0.05x²+ 10x + 5000 − 6500 x − 100

Note that

0.05x²+ 10x + 5000 − 6500

x − 100 = 1

20(x + 300) so

x→100lim

C(x) − C(100)

x − 100 = lim

x→100

1

20(x + 300) = 20

2