Math 1105(11)
Solutions to Assignment #1
1. (a) x 2 5 x 6 0
Since x 2 5 x 6 ( x 1)( x 6) 0 , we have
x 1 0 & x 6 0
Case 1:
or x 1
& x 6
x 1 0 & x 6 0
Case 2:
or x 1
& x 6 -------------- no solution
So the solution set is the closed interval [-6, 1].
(b) | 2 x 5 | 1
| 2 x 5 | 1 is equivalent to 1 < 2x + 5 < 1. From the first inequality we
have 6 < 2x or x > -3. From the second inequality we have 2x < -4 or x <
-2. So the solution set is (-3, -2).
2. Find the domain and range of each function.
(a) f ( x ) 3 2 x
Domain: {x | x 2, x
R}, Range: (-, 3].
f(x)
1
2 x3
Domain: x + 3 0 and
2 x 3 0 , which is [3, 1) (1, ).
Range: (-, 0) [1/2, ).
(b) f ( x )
x
1
-3
1(b)
3. Sketch the graphs of the functions:
(a) f ( x) ( x 1) 2 1 (b) f ( x) 1 | x 2 | .
f(x)
f(x)
2
1
1
1
2(a)
2(b)
4. Construct the following composite functions and specify the domain of each
(i) f f (x) (ii) f g (x) (iii) g f (x) (iv) g g (x)
a. f(x) = 1/x, g(x) = x/(2-x).
The domain of f(x) is {x | x 0 }. The domain of g(x) is {x | x 2 }.
1
x. Domain: { x | x 0 }
1/ x
1
2 x
x
f g ( x) f
.
x
(ii)
Domain: { x | x 0, x 2 }.
x
2 x
2 x
1
1
x x 1 x 1 .
(iii) g f ( x ) g (1 / x)
1
2x 1 x 2x 1 2x 1
2
x
x
(i) f f ( x) f (1 / x)
Domain: { x | x 0, x 1/2 }
x
x
x
2 x 2 x x 2 x x .
(iv) g g ( x) g
x
4 2 x x 2 x 4 3x 4 3x
2 x
2
2x
2 x
Domain: { x | x 2, x 4/3 }
(b)
f ( x)
2 x, g ( x ) 1 x
The domain of f(x) is {x | x 2 }. The domain of g(x) is all real number.
(i) f f ( x) f ( 2 x ) 2 2 x . Domain: [-2, 2].
(ii) f g ( x) f (1 x) 2 (1 x) 1 x . Domain: [-1, ).
(iii) g f ( x) g ( 2 x ) 1 2 x . Domain: (-, 2].
(iv) g g ( x) g (1 x) 1 (1 x) x . Domain: (-, ).
5. (a) If f ( x) x 4 and f g ( x) x , what is g (x) . Verify your answer.
g(x) = x 4 because f g ( x ) f ( x 4) ( x 4) 4 x.
(b) If g ( x) x 1 and f g ( x) 1 / x 2 , what is f (x ) . Verify your answer.
f ( x)
1
1
1
f g ( x) f ( x 1)
2
2 because
2
( x 1)
[( x 1) 1]
x
6. The motion of a particle moving along the x-axis so that at time t it is at
position x 3t 2 2t 1 m.
(a) Find the average velocity of the particle over the time intervals [1, 2] and
[2, 3].
3t 22 2t 2 1 (3t12 2t1 1)
Average velocity in [t1, t2] =
t 2 t1
3 (2) 2 2 2 1 (3 12 2 1 1)
11 m/s
2 1
3 (3) 2 2 3 1 (3 2 2 2 2 1)
17 m/s
Average velocity in [2, 3] =
32
Average velocity in [1, 2] =
(b) Make a table giving the average velocities of the particle over time
interval [1, 1+h], for h = 1, 0.1, 0.01, 0.001, 0.0001 second.
x 3 (1 h) 2 2(1 h) 1 (3 12 2 1 1)
8 3h
t
h
h
1
0.1
0.01
0.001
0.0001
x /t
11
8.3
8.03
8.003
8.0003
(c) Using the results of (b) to find the velocity of the particle at t=1 and t=2.
3 (t h) 2 2(t h) 1 (3t 2 2t 1)
Velocity at t = lim
h 0
h
3(t 2 2th h 2 ) 2(t h) 1 3t 2 2t 1
h0
h
h(6t 3h 2)
lim
6t 2
h0
h
lim
The velocity of the particle is 8 m/s at t=1 and 14 m/s at t=2.
7. Evaluate the limit or explain why it does not exist.
t2
( 4) 2
16
2
4 t 4 ( 4)
8
(a) tlim
4
x2 4
( x 2)( x 2)
lim
lim ( x 2) 2 2 4
x 2 x 2
x 2
x 2
x2
(b) lim
(c) lim
h 0
h 2h 2
h(1 2h)
1 2h
lim 2
lim
. No limit.
h 0 h (1 3h)
h 0 h (1 3h )
h 2 3h 3
2
(3 x) 2
|3 x |
(d) lim 9 6 x x lim
. No limit because
lim
x 3
x 3 x 3
x 3
x3
|3 x |
(3 x )
|3 x |
3 x
lim
lim
1 and lim
lim
1
x 3 x 3
x 3 x 3
x 3 x 3
x 3
x3
x 3
(e) lim
x0
| 2x 1 | | 2x 1 |
(2 x 1) (2 x 1)
4x
lim
lim
4
x 0
x0
x
x
x
( x 2)
| x2 |
| x2|
lim
1 and
doesnt exist because lim
x2 x2
x2
x2 x 2
x2
| x2|
lim
lim x 2 1 .
x 2 x 2
x 2 x 2
(f) lim
1
4
x2
4
x2
lim
2
2
lim
x
2
x
2
( x 2)( x 2)
x 2 x 4
( x 2)( x 2) x 4
(g) lim
x 2
lim
x 2
1
1
.
x2 4
8. Find the indicated one-side limit or explain why it does not exist.
(a) lim 2 x . Undefined because the domain of 2 x is x 2.
x 2
2 x 2 (2) 2 . The domain the same as that in (a)
(b) xlim
2
x 4 x 2 lim x 2 ( x 2 1) . Since the domain is x-1 or x1, the
(c) xlim
0
x 0
limit is undefined.
x2
if
2
9. Define f ( x ) 2 x 1 if
( x ) 2 if
x 1
1 x 0
x0
Find the indicated limits.
f ( x) lim ( x 2) 3 (b)
(a) xlim
1
x 1
lim f ( x ) lim ( 2 x 2 1) 3
x 1
x 1
f ( x ) lim ( 2 x 2 1) 1 (d) lim f ( x) lim ( x ) 2 2
(c) xlim
0
x 0
x 0
x 0
