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
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,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
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, 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
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, 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.
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manner. This document may not be copied, scanned, duplicated, forwarded, distributed, or posted on a website, in whole or part.
