Education.com
Try
Brainzy
Try
Plus

Calculus Basics Practice Test (page 2)

By — McGraw-Hill Professional
Updated on Aug 31, 2011

Answers:

  1.  

    SOLUTIONS TO EXERCISES Chapter 1

    SOLUTIONS TO EXERCISES Chapter 1

  2. In Fig. S1.2 , set A = 3.4, B = −π/2, C = 2π, SOLUTIONS TO EXERCISES Chapter 1 , SOLUTIONS TO EXERCISES Chapter 1 , F = 9/2, G = −29/10.

     

    SOLUTIONS TO EXERCISES Chapter 1

    Fig. S1.2

  3.  

    SOLUTIONS TO EXERCISES Chapter 1

    Fig. S1.3(a)

    SOLUTIONS TO EXERCISES Chapter 1

    Fig. S1.3(b)

    SOLUTIONS TO EXERCISES Chapter 1

    Fig. S1.3(c)

    SOLUTIONS TO EXERCISES Chapter 1

    Fig. S1.3(d)

    SOLUTIONS TO EXERCISES Chapter 1

    Fig. S1.3(e)

    SOLUTIONS TO EXERCISES Chapter 1

    Fig. S1.3(f)

  4. Let A = (2, −4), B = (−6, 3), C = (π, π 2 ), SOLUTIONS TO EXERCISES Chapter 1 , F = (1/3, −19/4).

    SOLUTIONS TO EXERCISES Chapter 1

  5.  

    SOLUTIONS TO EXERCISES Chapter 1

    Fig. S1.5(a)

    SOLUTIONS TO EXERCISES Chapter 1

    Fig. S1.5(b)

     

    SOLUTIONS TO EXERCISES Chapter 1

    Fig. S1.5(c)

    SOLUTIONS TO EXERCISES Chapter 1

    Fig. S1.5(d)

    SOLUTIONS TO EXERCISES Chapter 1

    Fig. S1.5(e)

    SOLUTIONS TO EXERCISES Chapter 1

    Fig. S1.5(f)

  6.  

    SOLUTIONS TO EXERCISES Chapter 1

    Fig. S1.6(a)

    SOLUTIONS TO EXERCISES Chapter 1

    Fig. S1.6(b)

    SOLUTIONS TO EXERCISES Chapter 1

    Fig. S1.6(c)

     

    SOLUTIONS TO EXERCISES Chapter 1

    Fig. S1.6(d)

    SOLUTIONS TO EXERCISES Chapter 1

    Fig. S1.6(e)

    SOLUTIONS TO EXERCISES Chapter 1

    Fig. S1.6(f)

  7.  

    SOLUTIONS TO EXERCISES Chapter 1

  8. (a) Slope is −3/8 hence line is y − (−9) = (−3/8) · ( x − 4)

    (b) Slope is 1 hence line is y − (−8) = 1 · ( x − (−4))

    (c) y − 6 = (−8) ( x − 4)

    (d) Slope is SOLUTIONS TO EXERCISES Chapter 1 hence line is y − 3 = (−1/8)( x − 2)

    (e) y = 6 x

    (f) Slope is −3 hence line is y − 7 = (−3)( x −(−4))

  9.  

    SOLUTIONS TO EXERCISES Chapter 1

    SOLUTIONS TO EXERCISES Chapter 1

     

    SOLUTIONS TO EXERCISES Chapter 1

  10. (a) Each person has one and only one father. This is a function.

    (b) Some men have more than one dog, others have none. This is not a function.

    (c) Some real numbers have two square roots while others have none. This is not a function.

    (d) Each positive integer has one and only one cube. This is a function.

    (e) Some cars have several drivers. In a one-car family, everyone drives the same car. So this is not a function.

    (f) Each toe is attached to one and only one foot. This is a function.

    (g) Each rational number succeeds one and only one integer. This is a function.

    (h) Each integer has one and only one successor. This is a function.

    (i) Each real number has a well defined square, and adding six is a well defined operation. This is a function.

  11.  

    SOLUTIONS TO EXERCISES Chapter 1

     

    SOLUTIONS TO EXERCISES Chapter 1

    SOLUTIONS TO EXERCISES Chapter 1

  12.  

    SOLUTIONS TO EXERCISES Chapter 1

  13. We check the first six identities.

    SOLUTIONS TO EXERCISES Chapter 1

    SOLUTIONS TO EXERCISES Chapter 1

  14. We shall do (a), (c), (e).

    SOLUTIONS TO EXERCISES Chapter 1

    Fig. S1.14(a)

    SOLUTIONS TO EXERCISES Chapter 1

    Fig. S1.14(c)

    SOLUTIONS TO EXERCISES Chapter 1

    Fig. S1.14(e)

  15. (a) θ = (15/2)°

    (b) θ = −60°

    (c) θ = 405°

    (d) θ = (405/4)°

    (e) θ = (540/π)°

    (f) θ = (−900/π)°

  16. (a) θ = 13π/36 radians

    (b) θ = π/18 radians

    (c) θ = −5π/12 radians

    (d) θ = −2π/3 radians

    (e) θ = π 2 /180 radians

    (e) θ = 157π/9000 radians

  17. (a) f ο g ( x ) = [( x − 1) 2 ] 2 + 2[( x − 1) 2 ] + 3; g ο f ( x ) = ([ x 2 + 2 x + 3] − 1) 2 .

    (b) SOLUTIONS TO EXERCISES Chapter 1

    (c) f ο g ( x ) = sin([cos( x 2x )] + 3[cos( x 2x )] 2 ); g ο f ( x ) = cos([sin( x + 3 x 2 )] 2 − [sin( x + 3 x 2 ]).

    (d) f ο g ( x ) = e ln( x − 5)+2 ; g ο f ( x ) = ln( e x +2 − 5).

    (e) f ο g ( x ) = sin([ln( x 2x )] 2 + [ln( x 2x )]); g ο f ( x ) = ln([sin( x 2 + x )] 2 − [sin( x 2 + x )]).

    (f) f ο g ( x ) = e [ e x 2 ] 2 ; f ο g ( x ) = e −[ e x 2 ] 2 .

    (g) f ο g ( x ) = [(2 x − 3)( x + 4)] · [(2 x − 3)( x + 4) + 1] · [(2 x − 3)( x + 4) + 2]; g ο f ( x ) = (2[ x ( x + 1)( x + 2)]−3)([ x ( x + 1)( x + 2)] + 4).

  18. (a) f is invertible, with f −1 ( t ) = ( t − 5) 1/3 .

    (b) g is not invertible since g (0) = g (1) = 0.

    (c) h is invertible, with h −1 ( t ) = sgn t · t 2 .

    (d) f is invertible, with f −1 ( t ) = ( t − 8) 1/5 .

    (e) g is invertible, with g −1 ( t ) = −[ln t ]/3.

    (f) h is not invertible, since SOLUTIONS TO EXERCISES Chapter 1 .

    (g) f is not invertible, since tan π/4 = tan 9π/4 = 1.

    (h) g is invertible, with SOLUTIONS TO EXERCISES Chapter 1 .

  19. We will do (a), (c), (e), and (g).

    SOLUTIONS TO EXERCISES Chapter 1

    Fig. S1.19(a)

     

    SOLUTIONS TO EXERCISES Chapter 1

    Fig. S1.19(c)

    SOLUTIONS TO EXERCISES Chapter 1

    Fig. S1.19(e)

    (g) Not invertible.

  20.  

    SOLUTIONS TO EXERCISES Chapter 1

View Full Article
Add your own comment

Ask a Question

Have questions about this article or topic? Ask
Ask
150 Characters allowed