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Solution Manual For Applied Calculus for Business, Economics, and the Social and Life Sciences 11th Edition By Laurence Hoffmann, Gerald Bradley, David Sobecki, Michael

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Chapter 1
Functions, Graphs, and Limits
1.1 Functions
1. f(x) = 3x + 5,
7. h(t ) = t 2 + 2t + 4,
f(0) = 3(0) + 5 = 5,
f(−1) = 3(−1) + 5 = 2, h(2)= 22 + 2(2) + 4= 2 3,
f(2) = 3(2) + 5 = 11.
h(0)= 02 + 2(0) + 4= 2,
2. f(x) = −7x + 1 h(−4) = (−4) 2 + 2(−4) + 4 = 2 3.
f(0) = −7(0) + 1 = 1
f(1) = −7(1) + 1 = −6
8. g (u=
) (u + 1)3 2
f(−2) = −7(−2) + 1 = 15
g (0) =
(0 + 1)3 2 =
1
3. f ( x) = 3 x 2 + 5 x − 2, g (−1) = (−1 + 1)3 2 = 0

( )
3
f (0) =3(0) 2 + 5(0) − 2 =−2, g (8) =
(8 + 1)3 2 = 9 =
27
f (−2) = 3(−2) 2 + 5(−2) − 2 = 0,
= 3(1) 2 + 5(1) −= 1
(2t − 1) −3/2 =
f (1) 2 6.
9. f (t ) = ,
( )
3
2t − 1
4. h(t ) =(2t + 1)3 h(−1) =(−2 + 1)3 =−1
1
h(0) =(0 + 1)3 =1 h(1) =(2 + 1)3 =27 =f (1) = 1,
3
 2(1) − 1 
 
1
5. g ( x)= x + , 1 1 1
x =
f (5) = 3
= 3
,
1  2(5) − 1   9 27
g (−1) =−1 + =−2,    
−1 1 1 1
1 =
f (13) = = .
g (1) =1 + = 2, 3 3 125
1  2(13) − 1   25 
   
1 5
g (2) = 2 + = .
2 2 1
10. f (t ) =
3 − 2t
6. 1
=f (1) = 1
3 − 2(1)
1 1
=f (−3) =
3 − 2(−3) 3
1 1
=f (0) =
3 − 2(0) 3

1
© 2013 by McGraw-Hill Education. This is proprietary material solely for authorized instructor use. Not authorized for sale or distribution in any
manner. This document may not be copied, scanned, duplicated, forwarded, distributed, or posted on a website, in whole or part.

,2 Chapter 1. Functions, Graphs, and Limits

11. f ( x) = x − x − 2 , of h=
(t ) t 2 + 1 consists of all real
f (1) = 1 − 1 − 2 = 1 − −1 = 1 − 1 = 0, numbers.
f (2) = 2 − 2 − 2 = 2 − 0 = 2,
f (3) = 3 − 3 − 2 = 3 − 1 = 3 − 2 = 2. x2 + 5
19. g ( x) =
x+2
12. g ( x)= 4 + x Since the denominator cannot be 0, the
g (−2) = 4 + −2 = 6 domain consists of all real numbers such
that x ≠ −2.
g (0) =4 + 0 =4
g (2) =4 + 2 =6 20. f ( x) = x3 − 3 x 2 + 2 x + 5
The domain consists of all real numbers.
−2 x + 4 if x ≤1
13. h( x) =  2
 x + 1 if x >1 21. f =
( x) 2x + 6
Since negative numbers do not have real
= (3)2 +=
h(3) 1 10 square roots, the domain is all real
h(1) =−2(1) + 4 =2 numbers such that 2x + 6 ≥ 0, or x ≥ −3.
h(0) =−2(0) + 4 =4
h(−3) =−2(−3) + 4 =10 t +1
22. f (t ) =
t2 − t − 2
3 if t < −5 t − t − 2 = (t − 2)(t + 1) ≠ 0
2

14. f (t ) = t + 1 if −5 ≤ t ≤ 5 if t ≠ −1 and t ≠ 2.

 t if t >5
t+2
f(−6) = 3 23. f (t ) =
f(−5) = −5 + 1 = −4 9 − t2
= =
f (16) 16 4 Since negative numbers do not have real
square roots and denominators cannot be
x zero, the domain is the set of all real
15. g ( x) = numbers such that 9 − t 2 > 0, namely
1 + x2
−3 < t < 3.
Since 1 + x 2 ≠ 0 for any real number, the
domain is the set of all real numbers.
24. h=
(s) s 2 − 4 is defined only if
16. Since x 2 − 1 = 0 for x = ±1 , f(x) is defined s 2 − 4 ≥ 0 or equivalently
only for x ≠ ±1 and the domain does not ( s − 2)( s + 2) ≥ 0 . This occurs when the
consist of the real numbers. factors ( s − 2) and ( s + 2) are zero or have
the same sign. This happens when s ≥ 2 or
17. f (t=
) 1− t
s ≤ −2 and these values of s form the
Since negative numbers do not have real domain of h.
square roots, the domain is all real
numbers such that 1 − t ≥ 0, or t ≤ 1. 25. f (u ) = 3u 2 + 2u − 6 and g(x) = x + 2, so
Therefore, the domain is not the set of all
f ( g (=
x)) f ( x + 2)
real numbers.
= 3( x + 2) 2 + 2( x + 2) − 6
18. The square root function only makes sense = 3 x 2 + 14 x + 10.
for non-negative numbers. Since
t 2 + 1 ≥ 0 for all real numbers t the domain

© 2013 by McGraw-Hill Education. This is proprietary material solely for authorized instructor use. Not authorized for sale or distribution in any
manner. This document may not be copied, scanned, duplicated, forwarded, distributed, or posted on a website, in whole or part.

, Chapter 1. Functions, Graphs, and Limits 3

26. f (u=
) u2 + 4 34. For f ( x=) 2x + 3 ,
f ( x − 1) = ( x − 1) 2 + 4 = x 2 − 2 x + 5 f ( x + h) − f ( x) (2( x + h) + 3) − (2 x + 3)
=
h h
2 x + 2h + 3 − 2 x − 3
27. f (u ) =(u − 1)3 + 2u 2 and g(x = x + 1, so =
h
f ( g (=
x)) f ( x + 1) 2h
= [( x + 1) − 1]3 + 2( x + 1) 2 =
h
= x3 + 2 x 2 + 4 x + 2. =2

28. f (=
u ) (2u + 10) 2
f ( x − 5)= [ 2( x − 5) + 10]2
= (2 x − 10 + 10) 2 = 4 x 2

1
29. f (u ) = and g(x) = x − 1, so 35. f ( x=
) 4 x − x2
u2
1 f ( x + h) − f ( x )
f ( g ( x))= f ( x − 1)= . h
( x − 1) 2
4( x + h) − ( x + h) 2 − (4 x − x 2 )
=
1 h
30. f (u ) =
u 4 x + 4h − ( x 2 + 2 xh + h 2 ) − 4 x + x 2
=
1 h
f ( x 2 + x − 2) =2
x + x−2 4 x + 4h − x − 2 xh − h 2 − 4 x + x 2
2
=
h
31. f (u=
) u + 1 and g ( x=
) x 2 − 1, so 4h − 2 xh − h 2
=
f ( g=
( x)) f ( x 2 − 1) h
h(4 − 2 x − h)
= ( x 2 − 1) + 1 =
h
= x2 =4 − 2 x − h
= x.
36. f ( x) = x 2
 1  1 f ( x + h) − f ( x ) ( x + h) 2 − x 2
=
32. f (u ) u=
2
, f  =
 x − 1  ( x − 1) 2 h h
x + 2 xh + h 2 − x 2
2
33. f(x) = 4 − 5x =
h
f ( x + h) − f ( x) 4 − 5( x + h) − (4 − 5 x) h(2 x + h)
= =
h h h
4 − 5 x − 5h − 4 + 5 x −5h = 2x + h
= = −5
h h

© 2013 by McGraw-Hill Education. This is proprietary material solely for authorized instructor use. Not authorized for sale or distribution in any
manner. This document may not be copied, scanned, duplicated, forwarded, distributed, or posted on a website, in whole or part.

, 4 Chapter 1. Functions, Graphs, and Limits

x
37. f ( x) = x = 4 x2
x +1 =0 4 x2 − x
f ( x + h) − f ( x ) =0 x(4 x − 1)
h 1
x+h − x =x 0,= x
( x + h ) +1 x +1 4
=
h Since squaring both sides can introduce
x+h − x extraneous solutions, one needs to check
( x + h + 1)( x + 1)
= x + h +1 x +1 ⋅ these values.
h ( x + h + 1)( x + 1)
( x + h)( x + 1) − x( x + h + 1)
=
h( x + 1)( x + h + 1)
x 2 + hx + x + h − x 2 − xh − x
=
h( x + 1)( x + h + 1)
h
=
h( x + 1)( x + h + 1)
1
=
( x + 1)( x + h + 1)

1 Also check remaining value to see if it is
38. f ( x) = in the domain of f and g functions. Since
x
f(0) and g(0) are both defined,
f ( x + h) − f ( x )
1
x+h
− 1x x( x + h) f(g(x)) = g(f(x)) when x = 0.
= ⋅
h h x ( x + h)
x − ( x + h) 40. f ( x) =
x 2 + 1, g ( x) =
1− x
=
hx( x + h) f ( g ( x)) =(1 − x)2 + 1
x−x−h
= = x 2 − 2 x + 2 and
hx( x + h)
g ( f ( x)) =
1 − ( x 2 + 1)
−h
= = − x2
hx( x + h)
So f ( g ( x)) = g ( f ( x)) means
−1
=
x ( x + h) x 2 − 2 x + 2 =− x 2
2 x2 − 2 x + 2 =0
39. f ( g ( x)) = f (1 − 3 x) = 1 − 3 x x2 − x + 1 =0
g ( f ( x)) = g ( x )= 1− 3 x but, by the quadratic formula, this last
equation has no solutions.
To solve 1 − 3 x =− 1 3 x , square both
sides, so
41. =
f ( g ( x)) f= =
x +3
 x + 3  2 x −2 + 3 ( )
1 − 3 x =−
1 6 x + 9x   x +3 − 1
x
 x−2 x −2
−3 x = −6 x + 9 x 2 x +3 + 3
6 x = 12 x  2x + 3  x −1
g=
( f ( x)) g  =  =
2 x +3 − 2
x
x = 2x  x −1  x −1
Squaring both sides again, Answer will be all real numbers for which
f and g are defined. So, f(g(x)) = g(f(x)) for
all real numbers except x = 1 and
x = 2.

© 2013 by McGraw-Hill Education. This is proprietary material solely for authorized instructor use. Not authorized for sale or distribution in any
manner. This document may not be copied, scanned, duplicated, forwarded, distributed, or posted on a website, in whole or part.

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Exam (elaborations)

Solution Manual For Applied Calculus for Business, Economics, and the Social and Life Sciences 11th Edition By Laurence Hoffmann, Gerald Bradley, David Sobecki, Michael

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