Unformatted text preview: ) = 0; f (0) = 0 f (0) = f (4) = 0; f (3) = 0; [0, 4], c = 3 3. f (2) = f (4) = 0, f (x) = 2x − 6, 2c − 6 = 0, c = 3
4. f (0) = f (2) = 0, f (x) = 3x − 6x + 2, 3c − 6c + 2 = 0; c =
2 2 6± √ √
36 − 24
= 1 ± 3/3
6 5. f (π/2) = f (3π/2) = 0, f (x) = − sin x, − sin c = 0, c = π
x2 − 4x + 1 c2 − 4c + 1
,
= 0, c2 − 4c + 1 = 0
(x − 2)2
(c − 2)2
√
√
√
4 ± 16 − 4
= 2 ± 3, of which only c = 2 − 3 is in (−1, 1)
c=
2 6. f (−1) = f (1) = 0, f (x) = 7. f (0) = f (4) = 0, f (x) = 8. 1
1
1
1
− √ , − √ = 0, c = 1
2 2x2 2c f (1) = f (3) = 0, f (x) = − 2
4
2
4
+ 2 , − 3 + 2 = 0, −6 + 4c = 0, c = 3/2
3
x
3x
c
3c Exercise Set 6.5 9. 202 6
3
f (8) − f (0)
= = = f (1.54); c = 1.54
8−0
8
4 10. 42 − 12
= 3, c = 1
6 − (−4) 11. f (−4) = 12, f (6) = 42, f (x) = 2x + 1, 2c + 1 = 12. f (−1) = −6, f (2) = 6, f (x) = 3x2 + 1, 3c2 + 1 =
in (−1, 2) 13. f (4) − f (0)
= 1.19 = f (0.77)
4−0 6 − (−6)
= 4, c2 = 1, c = ±1 of which only c = 1 is
2 − (−1) 1
2−1
1√
1
,√
=
= , c + 1 = 3/2, c + 1 = 9/4, c = 5/4
f (0) = 1, f (3) = 2, f (x) = √
3−0
3
2 x+1 2 c+1 14. f (3) = 10/3, f (4) = 17/4, f (x) = 1 − 1/x2 , 1 − 1/c2 =
√
which only c = 2 3 is in (3, 4)
15. f (−5) = 0, f (3) = 4, f (x) = − √ √
17/4 − 10/3
= 11/12, c2 = 12, c = ±2 3 of
4−3 √
1
x
c
4−0
= , −2c = 25 − c2 ,
, −√
=
2
2
3 − (−5)
2
25 − x
25 − c √
4c2 = 25 − c2 , c2 = 5, c = − 5
√
√
(we reject c = 5 because it does not satisfy the equation −2c = 25 − c2 )
16. f (2) = 1, f (5) = 1/4, f (x) = −1/(x − 1)2 ,
c = −1 (reject), or c = 3
17. (a) 1
1
1/4 − 1
= − , (c − 1)2 = 4, c − 1 = ±2,
=
2
(c − 1)
5−2
4 f (−2) = f (1) = 0 (b) c = −1.29
6 2 1
2 (c)
18. x0 = −1, x1 = −1.5, x2 = −1.32, x3 = −1.290, x4 = −1.2885843
f (−2) − f (1)
−16 − 5
=
= 7 so y − 5 = 7(x − 1),
−2 − 1
−3
y = 7x − 2 (a) m = (b) f (x) = 3x2 + 4 = 7 has solutions x = ±1;
discard x = 1, so c = −1 (d) 5
2 1 (c) y − f (−1) = 7(x − (−1)) or y = 7x + 2
20 19. (a) f (x) = sec2 x, sec2 c = 0 has no solution 20. (a) f (−1) = 1, f (8) = 4, f (x) = (b) tan x is not continuous on [0, π ] 2 −1/3
x
3 4−1
1
2 −1/3
=
c
= , c1/3 = 2, c = 8 which is not in (−1, 8).
3
8 − (−1)
3
(b) x2/3 is not diﬀerentiable at x = 0, which is in (−1, 8). 203 21. Chapter 6 Two xintercepts of f determine two solutions a and b of f (x) = 0; by Rolle’s Theorem there
exists a point c between a and b such that f (c) = 0, i.e. c is an xintercept for f . (b) 22. (a) f (x) = sin x = 0 at x = nπ , and f (x) = cos x = 0 at x = nπ + π/2, which lies between nπ and
(n + 1)π , (n = 0, ±1, ±2, . . .) f (x1 ) − f (x0 )
is the average rate of change of y with respect to x on the interval [x0 , x1 ]. By the
x1 − x0
MeanValue Theorem there is a value c in (x0 , x1 ) such that the instantaneous rate of change f (c) =
f (x1 ) − f (x0 )
.
x1 − x0 23. Let s(t) be the position function of the automobile for 0 ≤ t ≤ 5, then by the MeanValue Theorem
there is at least one point c in (0, 5) where
s (c) = v (c) = [s(5) − s(0)]/(5 − 0) = 4/5 = 0.8 mi/min = 48 mi/h. 24. Let T (t) denote the temperature at time with t = 0 denoting 11 AM, then T (0) = 76 and T (12) = 52.
(a) By the MeanValue Theorem there is a value c between 0 and 12 such that
T (c) = [T (12) − T (0)]/(12 − 0) = (52 − 76)/(12) = −2◦ F/h. (b) Assume that T (t1 ) = 88◦ F where 0 < t1 < 12, then there is at least one point c in (t1 , 12) where
T (c) = [T (12) − T (t1 )]/(12 − t1 ) = (52 − 88)/(12 − t1 ) = −36/(12 − t1 ). But 12 − t1 < 12 so
T (c) < −36/12 = −3◦ F/h. 25. Let f (t) and g (t) denote the distances from the ﬁrst and second runners to the starting point, and let
h(t) = f (t) − g (t). Since they start (at t = 0) and ﬁnish (at t = t1 ) at the same time, h(0) = h(t1 ) = 0,
so by Rolle’s Theorem there is a time t2 for which h (t2 ) = 0, i.e. f (t2 ) = g (t2 ); so they have the
same velocity at time t2 . 26. f (x) = x6 − 2x2 + x satisﬁes f (0) = f (1) = 0, so by Rolle’s Theorem f (c) = 0 for some c in (0, 1). 27. (a) By the Constant Diﬀerence Theorem f (x) − g (x) = k for some k ; since f (x0 ) = g (x0 ), k = 0, so
f (x) = g (x) for all x. (b) Set f (x) = sin2 x + cos2 x, g (x) = 1; then f (x) = 2 sin x cos x − 2 cos x sin x = 0 = g (x). Since
f (0) = 1 = g (0), f (x) = g (x) for all x. (a) By the Constant Diﬀerence Theorem f (x) − g (x) = k for some k ; since f (x0 ) − g (x0 ) = c, k = c,
so f (x) − g (x) = c for all x. (b) Set f (x) = (x − 1)3 , g (x) = (x2 + 3)(x − 3). Then
f (x) = 3(x − 1)2 , g (x) = (x2 + 3) + 2x(x − 3) = 3x2 − 6x + 3 = 3(x2 − 2x + 1) = 3(x − 1)2 ,
so f (x) = g (x) and hence f (x) − g (x) = k . Expand f (x) and g (x) to get
h(x) = f (x) − g (x) = (x3 − 3x2 + 3x − 1) − (x3 − 3x2 + 3x − 9) = 8. (c) h(x) = x3 − 3x2 + 3x − 1 − (x3 − 3x2 + 3x − 9) = 8 28. 29. 30. f (y ) − f (x)
= f (c),
y−x
so f (x) − f (y ) = f (c)x − y  ≤ M x − y ; if x > y exchange x and y ;...
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This document was uploaded on 02/23/2014 for the course MANAGMENT 2201 at University of Michigan.
 Spring '14
 The Land

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