Page 296 - Calculus Demystified

Chapter 3

Fig. S3.7 283

9. We see that
x
h(x) =
2
x − 1
2
x + 1

h (x) =−
2
(x − 1) 2
2
2x(x + 3)

h (x) =
2
(x − 1) 3
We see that the function is undefined at ±1, decreasing everywhere, and
has an inflection point only at 0. The sketch is shown in Fig. S3.9.
10. We know that
4π 3
V = r .
3
Therefore
dV 4π 2 dr
= · 3r .
dt 3 dt
Using the values V = 36π, r = 3, dV/dt =−2, we find that
dr
2
−2 = 4π · 3 ·
dt
hence
dr 1
=− in. per sec.
dt 18π