Page 293 - Calculus Demystified
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8. (a) We know that Solutions to Exercises
1 1
[f −1 = .
] (1) =
f (0) 3
(b) We know that
1 1
[f −1 = .
] (1) =
f (3) 8
(c) We know that
1 1
[f −1 = .
] (1) =
f (2) π 2
(d) We know that
1 1
[f −1 = .
] (1) =
f (1) 40
Chapter 3
1.
Fig. S3.1
2. Figure S3.2 shows a schematic of the imbedded cylinder. We see that the
volume of the imbedded cylinder, as a function of height h,is
2
V (h) = π · h · (3 − h/2) .
Then we solve
2
0 = V (h) = π · 9 − 6h + 3h /4 .
