Unit 5 Exam Review - Additional Problems 32 Polar Equations Pdf

Effort It

1.1 Functions and Function Notation

1 .

  1. ⓐ yes
  2. ⓑ yeah. (Notation: If two players had been tied for, say, fourth identify, then the name would non take been a function of rank.)

6 .

y = f ( x ) = x three 2 y = f ( x ) = ten 3 2

9 .

  1. ⓐ yes, because each bank business relationship has a single residual at whatever given fourth dimension
  2. ⓑ no, because several depository financial institution account numbers may have the same residue
  3. ⓒ no, because the same output may correspond to more than one input.

10 .

  1. ⓐ Aye, letter grade is a function of percent form;
  2. ⓑ No, information technology is not ane-to-one. There are 100 unlike percent numbers we could get simply only about 5 possible letter grades, and so there cannot be only 1 percent number that corresponds to each alphabetic character form.

12 .

No, considering information technology does not laissez passer the horizontal line test.

one.2 Domain and Range

1 .

{ − five , 0 , 5 , 10 , 15 } { − v , 0 , v , 10 , xv }

iii .

( − ∞ , one 2 ) ∪ ( 1 2 , ∞ ) ( − ∞ , 1 ii ) ∪ ( 1 2 , ∞ )

4 .

[ − 5 2 , ∞ ) [ − v 2 , ∞ )

five .

  1. ⓐ values that are less than or equal to –two, or values that are greater than or equal to –1 and less than 3;
  2. ⓑ { x | x ≤ − ii or − 1 ≤ 10 < 3 } { x | 10 ≤ − ii or − i ≤ x < 3 } ;
  3. ⓒ ( − ∞ , − 2 ] ∪ [ − 1 , 3 ) ( − ∞ , − 2 ] ∪ [ − one , iii )

half dozen .

domain =[1950,2002] range = [47,000,000,89,000,000]

7 .

domain: ( − ∞ , ii ] ; ( − ∞ , 2 ] ; range: ( − ∞ , 0 ] ( − ∞ , 0 ]

1.3 Rates of Change and Behavior of Graphs

1 .

$ 2.84 − $ 2.31 five  years = $ 0.53 5  years = $ 0.106 $ 2.84 − $ ii.31 v  years = $ 0.53 5  years = $ 0.106 per year.

4 .

The local maximum appears to occur at ( − 1 , 28 ) , ( − i , 28 ) , and the local minimum occurs at ( 5 , − 80 ) . ( five , − eighty ) . The function is increasing on ( − ∞ , − 1 ) ∪ ( five , ∞ ) ( − ∞ , − 1 ) ∪ ( five , ∞ ) and decreasing on ( − ane , 5 ) . ( − 1 , 5 ) .

Graph of a polynomial with a local maximum at (-1, 28) and local minimum at (5, -80).

one.4 Composition of Functions

1 .

( f g ) ( x ) = f ( ten ) yard ( ten ) = ( x − ane ) ( 10 2 − 1 ) = ten 3 − x 2 − x + one ( f − g ) ( x ) = f ( x ) − thou ( x ) = ( x − 1 ) − ( ten 2 − 1 ) = x − x 2 ( f g ) ( x ) = f ( x ) g ( x ) = ( ten − 1 ) ( x 2 − 1 ) = 10 3 − x two − ten + 1 ( f − g ) ( ten ) = f ( x ) − g ( x ) = ( 10 − 1 ) − ( ten 2 − 1 ) = x − x 2

No, the functions are not the same.

2 .

A gravitational forcefulness is all the same a force, and so a ( Thou ( r ) ) a ( G ( r ) ) makes sense as the dispatch of a planet at a distance r from the Dominicus (due to gravity), but M ( a ( F ) ) G ( a ( F ) ) does non make sense.

3 .

f ( 1000 ( 1 ) ) = f ( 3 ) = 3 f ( chiliad ( i ) ) = f ( 3 ) = 3 and g ( f ( 4 ) ) = g ( 1 ) = 3 g ( f ( 4 ) ) = thou ( 1 ) = 3

4 .

g ( f ( 2 ) ) = chiliad ( 5 ) = 3 k ( f ( 2 ) ) = one thousand ( 5 ) = 3

six .

[ − 4 , 0 ) ∪ ( 0 , ∞ ) [ − 4 , 0 ) ∪ ( 0 , ∞ )

7 .

Possible answer:

chiliad ( x ) = 4 + x ii g ( x ) = 4 + 10 2
h ( 10 ) = 4 iii − x h ( 10 ) = 4 3 − x
f = h ∘ g f = h ∘ 1000

1.five Transformation of Functions

1 .

b ( t ) = h ( t ) + 10 = − 4.9 t two + 30 t + 10 b ( t ) = h ( t ) + 10 = − 4.9 t 2 + xxx t + 10

ii .

The graphs of f ( 10 ) f ( x ) and g ( x ) g ( x ) are shown below. The transformation is a horizontal shift. The function is shifted to the left by two units.

Graph of a square root function and a horizontally shift square foot function.

4 .

g ( 10 ) = 1 x - 1 + 1 g ( x ) = 1 x - ane + 1

6 .

  1. ⓐ

    m ( 10 ) = − f ( x ) g ( 10 ) = − f ( 10 )

    x x -ii 0 two iv
    g ( x ) m ( x ) − 5 − v − 10 − 10 − 15 − fifteen − 20 − 20
  2. ⓑ

    h ( x ) = f ( − x ) h ( x ) = f ( − x )

    ten x -2 0 2 4
    h ( 10 ) h ( x ) 15 10 5 unknown

vii .

Graph of x^2 and its reflections.

Notice: g ( x ) = f ( − x ) g ( x ) = f ( − ten ) looks the same as f ( x ) f ( ten ) .

9 .

ten x 2 4 6 8
thousand ( x ) g ( ten ) 9 12 fifteen 0

11 .

g ( x ) = f ( 1 3 x ) thousand ( ten ) = f ( 1 3 ten ) so using the foursquare root role we get g ( x ) = i 3 x yard ( x ) = 1 3 x

1.6 Absolute Value Functions

2 .

using the variable p p for passing, | p − 80 | ≤ 20 | p − 80 | ≤ 20

3 .

f ( x ) = − | x + 2 | + 3 f ( x ) = − | x + two | + three

5 .

f ( 0 ) = ane , f ( 0 ) = 1 , so the graph intersects the vertical axis at ( 0 , 1 ) . ( 0 , 1 ) . f ( x ) = 0 f ( x ) = 0 when x = − 5 ten = − v and 10 = 1 x = 1 so the graph intersects the horizontal axis at ( − 5 , 0 ) ( − five , 0 ) and ( 1 , 0 ) . ( i , 0 ) .

7 .

k ≤ i k ≤ 1 or k ≥ 7 ; chiliad ≥ 7 ; in interval notation, this would be ( − ∞ , 1 ] ∪ [ vii , ∞ ) ( − ∞ , 1 ] ∪ [ 7 , ∞ )

1.7 Changed Functions

4 .

The domain of function f − one f − 1 is ( − ∞ , − two ) ( − ∞ , − 2 ) and the range of part f − one f − 1 is ( 1 , ∞ ) . ( 1 , ∞ ) .

v .

  1. f ( 60 ) = fifty. f ( 60 ) = l. In 60 minutes, 50 miles are traveled.
  2. f − 1 ( lx ) = 70. f − 1 ( 60 ) = lxx. To travel threescore miles, it will have seventy minutes.

8 .

f − 1 ( x ) = ( 2 − ten ) 2 ; domain of f : [ 0 , ∞ ) ; domain of f − 1 : ( − ∞ , 2 ] f − 1 ( 10 ) = ( 2 − x ) two ; domain of f : [ 0 , ∞ ) ; domain of f − 1 : ( − ∞ , 2 ]

1.i Section Exercises

1 .

A relation is a set of ordered pairs. A function is a special kind of relation in which no two ordered pairs accept the same offset coordinate.

3 .

When a vertical line intersects the graph of a relation more than than one time, that indicates that for that input there is more than than 1 output. At any particular input value, there tin be only one output if the relation is to exist a part.

5 .

When a horizontal line intersects the graph of a function more than once, that indicates that for that output there is more than one input. A function is one-to-one if each output corresponds to just i input.

27 .

f ( − 3 ) = − eleven ; f ( − three ) = − 11 ;
f ( two ) = − one ; f ( 2 ) = − 1 ;
f ( − a ) = − 2 a − 5 ; f ( − a ) = − 2 a − 5 ;
− f ( a ) = − two a + 5 ; − f ( a ) = − two a + 5 ;
f ( a + h ) = ii a + 2 h − 5 f ( a + h ) = 2 a + two h − 5

29 .

f ( − three ) = 5 + 5 ; f ( − three ) = 5 + v ;
f ( 2 ) = 5 ; f ( 2 ) = 5 ;
f ( − a ) = two + a + 5 ; f ( − a ) = 2 + a + 5 ;
− f ( a ) = − 2 − a − 5 ; − f ( a ) = − ii − a − 5 ;
f ( a + h ) = ii − a − h + 5 f ( a + h ) = ii − a − h + 5

31 .

f ( − 3 ) = ii ; f ( − 3 ) = ii ; f ( 2 ) = ane − iii = − two ; f ( 2 ) = ane − 3 = − 2 ;
f ( − a ) = | − a − 1 | − | − a + 1 | ; f ( − a ) = | − a − ane | − | − a + 1 | ;
− f ( a ) = − | a − 1 | + | a + 1 | ; − f ( a ) = − | a − 1 | + | a + 1 | ;
f ( a + h ) = | a + h − 1 | − | a + h + ane | f ( a + h ) = | a + h − 1 | − | a + h + 1 |

33 .

g ( ten ) − one thousand ( a ) ten − a = x + a + 2 , x ≠ a g ( x ) − g ( a ) x − a = x + a + two , 10 ≠ a

35 .

  1. ⓐ f ( − two ) = fourteen ; f ( − ii ) = 14 ;
  2. ⓑ x = 3 ten = 3

37 .

  1. ⓐ f ( 5 ) = ten ; f ( 5 ) = ten ;
  2. ⓑ x = − 1 x = − 1 or x = iv x = 4

39 .

  1. ⓐ f ( t ) = 6 − two iii t ; f ( t ) = 6 − two 3 t ;
  2. ⓑ f ( − 3 ) = 8 ; f ( − three ) = viii ;
  3. ⓒ t = six t = 6

53 .

  1. ⓐ f ( 0 ) = 1 ; f ( 0 ) = 1 ;
  2. ⓑ f ( x ) = − 3 , x = − 2 f ( ten ) = − 3 , x = − 2 or x = 2 x = 2

55 .

not a function and then information technology is also non a one-to-1 role

59 .

role, but non ane-to-one

67 .

f ( x ) = 1 , 10 = 2 f ( x ) = ane , x = 2

69 .

f ( − ii ) = 14 ; f ( − 1 ) = xi ; f ( 0 ) = 8 ; f ( ane ) = 5 ; f ( 2 ) = two f ( − ii ) = 14 ; f ( − 1 ) = 11 ; f ( 0 ) = 8 ; f ( ane ) = 5 ; f ( 2 ) = 2

71 .

f ( − 2 ) = 4 ; f ( − ane ) = 4.414 ; f ( 0 ) = 4.732 ; f ( one ) = 5 ; f ( 2 ) = 5.236 f ( − 2 ) = 4 ; f ( − 1 ) = 4.414 ; f ( 0 ) = iv.732 ; f ( i ) = 5 ; f ( 2 ) = v.236

73 .

f ( − two ) = 1 9 ; f ( − one ) = 1 three ; f ( 0 ) = ane ; f ( 1 ) = 3 ; f ( 2 ) = 9 f ( − 2 ) = 1 ix ; f ( − i ) = 1 3 ; f ( 0 ) = ane ; f ( 1 ) = 3 ; f ( ii ) = 9

77 .

[ 0 ,  100 ] [ 0 ,  100 ]

Graph of a parabola.

79 .

[ − 0.001 ,  0 .001 ] [ − 0.001 ,  0 .001 ]

Graph of a parabola.

81 .

[ − 1 , 000 , 000 ,  one,000,000 ] [ − i , 000 , 000 ,  1,000,000 ]

Graph of a cubic function.

83 .

[ 0 ,  10 ] [ 0 ,  ten ]

Graph of a square root function.

85 .

[ −0.1 , 0.1 ] [ −0.i , 0.1 ]

Graph of a square root function.

87 .

[ − 100 ,  100 ] [ − 100 ,  100 ]

Graph of a cubic root function.

89 .

  1. ⓐ g ( 5000 ) = fifty ; m ( 5000 ) = 50 ;
  2. ⓑ The number of cubic yards of dirt required for a garden of 100 square anxiety is ane.

91 .

  1. ⓐ The acme of a rocket in a higher place basis after one second is 200 ft.
  2. ⓑ the meridian of a rocket above footing after 2 seconds is 350 ft.

one.ii Section Exercises

1 .

The domain of a office depends upon what values of the independent variable make the office undefined or imaginary.

3 .

There is no restriction on x x for f ( x ) = x iii f ( x ) = ten three because you can have the cube root of any real number. And then the domain is all real numbers, ( − ∞ , ∞ ) . ( − ∞ , ∞ ) . When dealing with the set of real numbers, you cannot take the square root of negative numbers. Then ten x -values are restricted for f ( x ) = ten f ( x ) = x to nonnegative numbers and the domain is [ 0 , ∞ ) . [ 0 , ∞ ) .

5 .

Graph each formula of the piecewise function over its corresponding domain. Utilize the same scale for the 10 10 -centrality and y y -axis for each graph. Indicate inclusive endpoints with a solid circle and exclusive endpoints with an open circle. Employ an arrow to signal − ∞ − ∞ or ∞ . ∞ . Combine the graphs to discover the graph of the piecewise part.

xv .

( − ∞ , − 1 two ) ∪ ( − 1 2 , ∞ ) ( − ∞ , − 1 ii ) ∪ ( − 1 two , ∞ )

17 .

( − ∞ , − 11 ) ∪ ( − 11 , ii ) ∪ ( ii , ∞ ) ( − ∞ , − 11 ) ∪ ( − 11 , 2 ) ∪ ( ii , ∞ )

xix .

( − ∞ , − iii ) ∪ ( − 3 , 5 ) ∪ ( 5 , ∞ ) ( − ∞ , − iii ) ∪ ( − 3 , 5 ) ∪ ( 5 , ∞ )

25 .

( − ∞ , − 9 ) ∪ ( − 9 , 9 ) ∪ ( 9 , ∞ ) ( − ∞ , − 9 ) ∪ ( − ix , 9 ) ∪ ( 9 , ∞ )

27 .

domain: ( 2 , 8 ] , ( 2 , 8 ] , range [ 6 , eight ) [ 6 , eight )

29 .

domain: [ − 4 ,  iv], [ − 4 ,  4], range: [ 0 ,  2] [ 0 ,  2]

31 .

domain: [ − 5 , 3 ) , [ − v , 3 ) , range: [ 0 , ii ] [ 0 , two ]

33 .

domain: ( − ∞ , 1 ] , ( − ∞ , one ] , range: [ 0 , ∞ ) [ 0 , ∞ )

35 .

domain: [ − 6 , − 1 6 ] ∪ [ 1 6 , six ] ; [ − half dozen , − 1 half dozen ] ∪ [ 1 six , 6 ] ; range: [ − 6 , − 1 half-dozen ] ∪ [ 1 six , vi ] [ − half-dozen , − 1 6 ] ∪ [ ane 6 , six ]

37 .

domain: [ − 3 , ∞ ) ; [ − three , ∞ ) ; range: [ 0 , ∞ ) [ 0 , ∞ )

39 .

domain: ( − ∞ , ∞ ) ( − ∞ , ∞ )

Graph of f(x).

41 .

domain: ( − ∞ , ∞ ) ( − ∞ , ∞ )

Graph of f(x).

43 .

domain: ( − ∞ , ∞ ) ( − ∞ , ∞ )

Graph of f(x).

45 .

domain: ( − ∞ , ∞ ) ( − ∞ , ∞ )

Graph of f(x).

47 .

f ( − three ) = 1 ; f ( − 2 ) = 0 ; f ( − 1 ) = 0 ; f ( 0 ) = 0 f ( − 3 ) = ane ; f ( − two ) = 0 ; f ( − 1 ) = 0 ; f ( 0 ) = 0

49 .

f ( − 1 ) = − 4 ; f ( 0 ) = 6 ; f ( 2 ) = 20 ; f ( iv ) = 34 f ( − 1 ) = − 4 ; f ( 0 ) = half dozen ; f ( 2 ) = 20 ; f ( 4 ) = 34

51 .

f ( − 1 ) = − v ; f ( 0 ) = three ; f ( ii ) = 3 ; f ( iv ) = 16 f ( − i ) = − v ; f ( 0 ) = iii ; f ( 2 ) = three ; f ( 4 ) = 16

53 .

domain: ( − ∞ , 1 ) ∪ ( 1 , ∞ ) ( − ∞ , ane ) ∪ ( 1 , ∞ )

55 .

Graph of the equation from [-0.5, -0.1].

window: [ − 0.five , − 0.1 ] ; [ − 0.5 , − 0.1 ] ; range: [ 4 , 100 ] [ 4 , 100 ]

Graph of the equation from [0.1, 0.5].

window: [ 0.1 , 0.v ] ; [ 0.1 , 0.five ] ; range: [ 4 , 100 ] [ 4 , 100 ]

59 .

Many answers. 1 function is f ( x ) = 1 ten − 2 . f ( ten ) = 1 x − 2 .

one.3 Section Exercises

1 .

Yeah, the average rate of change of all linear functions is constant.

3 .

The absolute maximum and minimum relate to the unabridged graph, whereas the local extrema relate only to a specific region around an open up interval.

xi .

− 1 xiii ( thirteen + h ) − one thirteen ( 13 + h )

13 .

3 h 2 + nine h + 9 3 h ii + 9 h + 9

xix .

increasing on ( − ∞ , − 2.5 ) ∪ ( ane , ∞ ) , ( − ∞ , − 2.v ) ∪ ( one , ∞ ) , decreasing on ( − two.5 , i ) ( − two.5 , 1 )

21 .

increasing on ( − ∞ , i ) ∪ ( 3 , 4 ) , ( − ∞ , ane ) ∪ ( 3 , iv ) , decreasing on ( ane , 3 ) ∪ ( iv , ∞ ) ( 1 , 3 ) ∪ ( iv , ∞ )

23 .

local maximum: ( − three , 50 ) , ( − three , fifty ) , local minimum: ( three , − l ) ( 3 , − fifty )

25 .

absolute maximum at approximately ( 7 , 150 ) , ( 7 , 150 ) , absolute minimum at approximately ( −vii.5 , −220 ) ( −7.v , −220 )

35 .

Local minimum at ( 3 , − 22 ) , ( 3 , − 22 ) , decreasing on ( − ∞ , three ) , ( − ∞ , 3 ) , increasing on ( 3 , ∞ ) ( iii , ∞ )

37 .

Local minimum at ( − 2 , − 2 ) , ( − two , − ii ) , decreasing on ( − 3 , − 2 ) , ( − 3 , − 2 ) , increasing on ( − two , ∞ ) ( − two , ∞ )

39 .

Local maximum at ( − 0.v , 6 ) , ( − 0.5 , 6 ) , local minima at ( − 3.25 , − 47 ) ( − three.25 , − 47 ) and ( ii.i , − 32 ) , ( 2.one , − 32 ) , decreasing on ( − ∞ , − 3.25 ) ( − ∞ , − three.25 ) and ( − 0.v , ii.1 ) , ( − 0.5 , 2.1 ) , increasing on ( − 3.25 , − 0.five ) ( − 3.25 , − 0.5 ) and ( 2.1 , ∞ ) ( 2.one , ∞ )

45 .

2.seven gallons per minute

47 .

approximately –0.6 milligrams per day

1.4 Section Exercises

ane .

Notice the numbers that brand the function in the denominator thou g equal to zip, and check for any other domain restrictions on f f and k , g , such as an fifty-fifty-indexed root or zeros in the denominator.

3 .

Yes. Sample answer: Let f ( x ) = x + 1  and yard ( x ) = x − 1. f ( x ) = x + 1  and g ( x ) = x − ane. Then f ( g ( x ) ) = f ( x − 1 ) = ( ten − 1 ) + 1 = x f ( thousand ( x ) ) = f ( x − 1 ) = ( ten − 1 ) + 1 = x and g ( f ( 10 ) ) = k ( x + 1 ) = ( 10 + i ) − 1 = x . m ( f ( x ) ) = m ( 10 + 1 ) = ( x + 1 ) − i = x . So f ∘ g = m ∘ f . f ∘ one thousand = g ∘ f .

5 .

( f + grand ) ( 10 ) = 2 x + 6 , ( f + 1000 ) ( x ) = 2 x + half-dozen , domain: ( − ∞ , ∞ ) ( − ∞ , ∞ )

( f − g ) ( x ) = 2 x 2 + ii x − 6 , ( f − g ) ( x ) = ii x 2 + 2 x − 6 , domain: ( − ∞ , ∞ ) ( − ∞ , ∞ )

( f g ) ( x ) = − x 4 − 2 x 3 + 6 10 2 + 12 x , ( f thousand ) ( x ) = − ten four − ii 10 3 + half-dozen x 2 + 12 ten , domain: ( − ∞ , ∞ ) ( − ∞ , ∞ )

( f g ) ( x ) = x 2 + 2 10 half-dozen − x 2 , ( f g ) ( x ) = x 2 + 2 x six − x 2 , domain: ( − ∞ , − 6 ) ∪ ( − vi , 6 ) ∪ ( 6 , ∞ ) ( − ∞ , − half dozen ) ∪ ( − 6 , 6 ) ∪ ( 6 , ∞ )

7 .

( f + thousand ) ( x ) = 4 x 3 + 8 ten 2 + 1 2 ten , ( f + g ) ( x ) = 4 x 3 + 8 ten 2 + ane ii x , domain: ( − ∞ , 0 ) ∪ ( 0 , ∞ ) ( − ∞ , 0 ) ∪ ( 0 , ∞ )

( f − 1000 ) ( 10 ) = 4 x three + 8 10 ii − 1 2 x , ( f − g ) ( x ) = 4 x three + 8 x 2 − i 2 x , domain: ( − ∞ , 0 ) ∪ ( 0 , ∞ ) ( − ∞ , 0 ) ∪ ( 0 , ∞ )

( f g ) ( x ) = x + two , ( f g ) ( ten ) = x + 2 , domain: ( − ∞ , 0 ) ∪ ( 0 , ∞ ) ( − ∞ , 0 ) ∪ ( 0 , ∞ )

( f thousand ) ( x ) = four x 3 + 8 x 2 , ( f 1000 ) ( x ) = 4 x 3 + 8 x 2 , domain: ( − ∞ , 0 ) ∪ ( 0 , ∞ ) ( − ∞ , 0 ) ∪ ( 0 , ∞ )

9 .

( f + 1000 ) ( x ) = three ten 2 + x − 5 , ( f + one thousand ) ( x ) = iii ten two + x − 5 , domain: [ 5 , ∞ ) [ 5 , ∞ )

( f − thousand ) ( ten ) = three ten 2 − ten − v , ( f − chiliad ) ( 10 ) = 3 x 2 − x − 5 , domain: [ five , ∞ ) [ v , ∞ )

( f g ) ( x ) = three x 2 10 − v , ( f yard ) ( x ) = 3 x 2 x − 5 , domain: [ 5 , ∞ ) [ 5 , ∞ )

( f g ) ( x ) = iii 10 2 x − 5 , ( f thousand ) ( x ) = 3 x two x − 5 , domain: ( 5 , ∞ ) ( five , ∞ )

xi .

  1. ⓐ 3
  2. ⓑ f ( g ( x ) ) = two ( three x − five ) 2 + 1 ; f ( chiliad ( ten ) ) = ii ( 3 x − 5 ) two + 1 ;
  3. ⓒ thousand ( f ) ( x ) ) = 6 x ii − 2 ; g ( f ) ( 10 ) ) = vi 10 2 − 2 ;
  4. ⓓ ( thou ∘ grand ) ( x ) = 3 ( iii x − v ) − five = 9 x − 20 ; ( k ∘ g ) ( x ) = 3 ( iii x − five ) − v = nine x − 20 ;
  5. ⓔ ( f ∘ f ) ( − two ) = 163 ( f ∘ f ) ( − 2 ) = 163

13 .

f ( g ( x ) ) = x 2 + 3 + 2 , g ( f ( x ) ) = x + 4 x + vii f ( g ( x ) ) = x 2 + 3 + 2 , g ( f ( 10 ) ) = x + 4 x + 7

xv .

f ( thou ( x ) ) = ten + 1 x 3 3 = 10 + ane 3 ten , k ( f ( ten ) ) = x 3 + ane 10 f ( 1000 ( 10 ) ) = x + 1 x three 3 = 10 + 1 3 x , g ( f ( x ) ) = x iii + one x

17 .

( f ∘ g ) ( x ) = 1 2 ten + 4 − 4 = ten two , ( g ∘ f ) ( x ) = 2 x − 4 ( f ∘ g ) ( ten ) = 1 two ten + 4 − iv = 10 two , ( g ∘ f ) ( 10 ) = 2 x − 4

nineteen .

f ( one thousand ( h ( x ) ) ) = ( one x + 3 ) 2 + 1 f ( thousand ( h ( ten ) ) ) = ( one x + 3 ) ii + one

21 .

  • ⓐ Text ( g ∘ f ) ( x ) = − 3 two − 4 x ; ( k ∘ f ) ( 10 ) = − iii 2 − 4 x ;
  • ⓑ ( − ∞ , 1 2 ) ( − ∞ , 1 2 )

23 .

  1. ⓐ ( 0 , 2 ) ∪ ( 2 , ∞ ) ; ( 0 , 2 ) ∪ ( 2 , ∞ ) ;
  2. ⓑ ( − ∞ , − 2 ) ∪ ( 2 , ∞ ) ; ( − ∞ , − 2 ) ∪ ( 2 , ∞ ) ; c. ( 0 , ∞ ) ( 0 , ∞ )

27 .

sample: f ( x ) = 10 iii 1000 ( x ) = x − 5 f ( x ) = 10 iii grand ( x ) = 10 − five

29 .

sample: f ( x ) = four x thou ( x ) = ( 10 + two ) ii f ( ten ) = iv x grand ( x ) = ( x + ii ) 2

31 .

sample: f ( x ) = x 3 g ( x ) = 1 2 x − 3 f ( x ) = x three 1000 ( x ) = i two x − 3

33 .

sample: f ( x ) = 10 4 yard ( x ) = iii 10 − 2 x + 5 f ( ten ) = x 4 chiliad ( x ) = iii x − 2 x + five

35 .

sample: f ( x ) = x f ( x ) = x
g ( ten ) = two ten + half-dozen g ( x ) = 2 x + 6

37 .

sample: f ( x ) = 10 three f ( 10 ) = ten 3
g ( x ) = ( ten − one ) grand ( x ) = ( ten − 1 )

39 .

sample: f ( 10 ) = x iii f ( 10 ) = x 3
1000 ( x ) = 1 ten − 2 m ( ten ) = 1 x − 2

41 .

sample: f ( x ) = x f ( x ) = x
g ( x ) = 2 x − i 3 x + 4 grand ( x ) = 2 ten − 1 3 ten + 4

73 .

f ( g ( 0 ) ) = 27 , g ( f ( 0 ) ) = − 94 f ( chiliad ( 0 ) ) = 27 , grand ( f ( 0 ) ) = − 94

75 .

f ( g ( 0 ) ) = 1 5 , chiliad ( f ( 0 ) ) = 5 f ( one thousand ( 0 ) ) = one five , chiliad ( f ( 0 ) ) = 5

77 .

xviii ten 2 + 60 10 + 51 18 x 2 + 60 10 + 51

79 .

thou ∘ g ( x ) = 9 10 + twenty grand ∘ g ( 10 ) = ix x + 20

87 .

( f ∘ g ) ( six ) = 6 ( f ∘ g ) ( half-dozen ) = 6 ; ( k ∘ f ) ( 6 ) = 6 ( chiliad ∘ f ) ( half dozen ) = 6

89 .

( f ∘ g ) ( 11 ) = xi , ( yard ∘ f ) ( eleven ) = 11 ( f ∘ grand ) ( eleven ) = 11 , ( m ∘ f ) ( 11 ) = 11

93 .

A ( t ) = π ( 25 t + 2 ) ii A ( t ) = π ( 25 t + 2 ) 2 and A ( 2 ) = π ( 25 iv ) 2 = 2500 π A ( 2 ) = π ( 25 iv ) two = 2500 π square inches

95 .

A ( five ) = π ( 2 ( 5 ) + one ) two = 121 π A ( 5 ) = π ( 2 ( 5 ) + 1 ) 2 = 121 π square units

97 .

  • ⓐ Due north ( T ( t ) ) = 23 ( 5 t + ane.5 ) 2 − 56 ( five t + 1.5 ) + 1 ; N ( T ( t ) ) = 23 ( 5 t + 1.5 ) two − 56 ( 5 t + i.5 ) + 1 ;
  • ⓑ 3.38 hours

1.v Section Exercises

i .

A horizontal shift results when a constant is added to or subtracted from the input. A vertical shifts results when a constant is added to or subtracted from the output.

iii .

A horizontal compression results when a constant greater than 1 is multiplied by the input. A vertical compression results when a constant between 0 and i is multiplied by the output.

5 .

For a function f , f , substitute ( − 10 ) ( − ten ) for ( x ) ( x ) in f ( x ) . f ( x ) . Simplify. If the resulting function is the same as the original role, f ( − x ) = f ( ten ) , f ( − ten ) = f ( x ) , then the function is even. If the resulting role is the reverse of the original role, f ( − x ) = − f ( x ) , f ( − x ) = − f ( ten ) , and then the original function is odd. If the function is not the same or the opposite, then the function is neither odd nor even.

seven .

1000 ( x ) = | x - 1 | − 3 g ( ten ) = | x - i | − 3

9 .

thou ( 10 ) = ane ( 10 + 4 ) two + 2 chiliad ( 10 ) = i ( x + 4 ) 2 + ii

11 .

The graph of f ( x + 43 ) f ( x + 43 ) is a horizontal shift to the left 43 units of the graph of f . f .

13 .

The graph of f ( 10 - 4 ) f ( x - 4 ) is a horizontal shift to the right 4 units of the graph of f . f .

15 .

The graph of f ( ten ) + 8 f ( x ) + 8 is a vertical shift up 8 units of the graph of f . f .

17 .

The graph of f ( x ) − 7 f ( x ) − 7 is a vertical shift down 7 units of the graph of f . f .

19 .

The graph of f ( x + 4 ) − i f ( x + four ) − 1 is a horizontal shift to the left four units and a vertical shift down ane unit of the graph of f . f .

21 .

decreasing on ( − ∞ , − three ) ( − ∞ , − iii ) and increasing on ( − 3 , ∞ ) ( − 3 , ∞ )

23 .

decreasing on [ 0 , ∞ ) [ 0 , ∞ )

31 .

1000 ( x ) = f ( ten - one ) , h ( 10 ) = f ( 10 ) + ane g ( 10 ) = f ( x - ane ) , h ( x ) = f ( x ) + 1

33 .

f ( 10 ) = | x - 3 | − 2 f ( x ) = | 10 - 3 | − two

35 .

f ( x ) = x + 3 − 1 f ( x ) = x + 3 − ane

37 .

f ( x ) = ( x - 2 ) 2 f ( x ) = ( x - ii ) 2

39 .

f ( x ) = | x + three | − 2 f ( 10 ) = | x + 3 | − two

43 .

f ( x ) = − ( x + 1 ) 2 + two f ( x ) = − ( x + 1 ) 2 + ii

45 .

f ( x ) = − x + 1 f ( ten ) = − x + 1

53 .

The graph of g g is a vertical reflection (across the x 10 -axis) of the graph of f . f .

55 .

The graph of g g is a vertical stretch past a factor of 4 of the graph of f . f .

57 .

The graph of g yard is a horizontal pinch past a factor of 1 five i 5 of the graph of f . f .

59 .

The graph of g 1000 is a horizontal stretch by a factor of 3 of the graph of f . f .

61 .

The graph of g one thousand is a horizontal reflection across the y y -axis and a vertical stretch by a factor of three of the graph of f . f .

63 .

k ( x ) = | − four x | g ( x ) = | − 4 ten |

65 .

1000 ( x ) = 1 3 ( 10 + 2 ) 2 − 3 thousand ( x ) = 1 3 ( x + 2 ) 2 − three

67 .

grand ( x ) = i 2 ( x - 5 ) 2 + i g ( x ) = 1 2 ( x - 5 ) 2 + i

69 .

The graph of the function f ( ten ) = x two f ( x ) = x 2 is shifted to the left 1 unit, stretched vertically by a cistron of 4, and shifted downwards 5 units.

Graph of a parabola.

71 .

The graph of f ( x ) = | x | f ( x ) = | x | is stretched vertically past a factor of ii, shifted horizontally iv units to the right, reflected across the horizontal axis, and then shifted vertically three units up.

Graph of an absolute function.

73 .

The graph of the office f ( x ) = 10 three f ( 10 ) = 10 3 is compressed vertically by a factor of ane 2 . i two .

Graph of a cubic function.

75 .

The graph of the function is stretched horizontally by a gene of 3 and then shifted vertically downwardly past 3 units.

Graph of a cubic function.

77 .

The graph of f ( x ) = x f ( x ) = x is shifted correct 4 units and and so reflected across the vertical line x = 4. x = four.

Graph of a square root function.

1.vi Section Exercises

one .

Isolate the absolute value term and so that the equation is of the form | A | = B . | A | = B . Form one equation past setting the expression inside the accented value symbol, A , A , equal to the expression on the other side of the equation, B . B . Form a second equation past setting A A equal to the opposite of the expression on the other side of the equation, − B . − B . Solve each equation for the variable.

3 .

The graph of the absolute value function does not cross the x x -axis, then the graph is either completely above or completely below the x x -centrality.

5 .

Starting time determine the purlieus points by finding the solution(s) of the equation. Employ the boundary points to form possible solution intervals. Cull a test value in each interval to determine which values satisfy the inequality.

seven .

| x + iv | = 1 ii | x + 4 | = ane 2

9 .

| f ( x ) − viii | < 0.03 | f ( x ) − eight | < 0.03

13 .

{ - 9 4 , 13 four } { - ix 4 , 13 4 }

fifteen .

{ 10 three , 20 3 } { 10 three , xx 3 }

17 .

{ eleven v , 29 five } { xi 5 , 29 5 }

19 .

{ 5 2 , seven 2 } { 5 two , seven 2 }

23 .

{ − 57 , 27 } { − 57 , 27 }

25 .

( 0 , − 8 ) ; ( − 6 , 0 ) , ( four , 0 ) ( 0 , − 8 ) ; ( − 6 , 0 ) , ( 4 , 0 )

27 .

( 0 , − 7 ) ; ( 0 , − vii ) ; no 10 x -intercepts

29 .

( − ∞ , − 8 ) ∪ ( 12 , ∞ ) ( − ∞ , − 8 ) ∪ ( 12 , ∞ )

33 .

( − ∞ , − 8 three ] ∪ [ six , ∞ ) ( − ∞ , − 8 3 ] ∪ [ 6 , ∞ )

35 .

( − ∞ , − 8 3 ] ∪ [ 16 , ∞ ) ( − ∞ , − 8 three ] ∪ [ 16 , ∞ )

53 .

range: [ 0 , xx ] [ 0 , 20 ]

Graph of an absolute function.

55 .

x - x - intercepts:

Graph of an absolute function.

59 .

At that place is no solution for a a that will keep the function from having a y y -intercept. The absolute value function always crosses the y y -intercept when x = 0. x = 0.

61 .

| p − 0.08 | ≤ 0.015 | p − 0.08 | ≤ 0.015

63 .

| x − v.0 | ≤ 0.01 | x − v.0 | ≤ 0.01

i.7 Section Exercises

i .

Each output of a function must have exactly ane output for the function to exist 1-to-one. If whatever horizontal line crosses the graph of a function more than once, that means that y y -values echo and the function is not ane-to-one. If no horizontal line crosses the graph of the function more than one time, so no y y -values echo and the function is one-to-1.

3 .

Yep. For example, f ( x ) = 1 x f ( x ) = 1 x is its own inverse.

5 .

Given a role y = f ( ten ) , y = f ( 10 ) , solve for ten 10 in terms of y . y . Interchange the ten ten and y . y . Solve the new equation for y . y . The expression for y y is the inverse, y = f − i ( x ) . y = f − ane ( x ) .

vii .

f − 1 ( x ) = 10 − iii f − 1 ( x ) = x − 3

9 .

f − ane ( ten ) = ii − x f − 1 ( x ) = ii − x

11 .

f − 1 ( x ) = − two x x − 1 f − one ( x ) = − 2 ten x − 1

13 .

domain of f ( 10 ) : [ − 7 , ∞ ) ; f − ane ( x ) = x − 7 f ( x ) : [ − 7 , ∞ ) ; f − 1 ( x ) = x − seven

15 .

domain of f ( x ) : [ 0 , ∞ ) ; f − i ( ten ) = ten + 5 f ( 10 ) : [ 0 , ∞ ) ; f − 1 ( ten ) = ten + 5

xvi .

  • ⓐ f ( thousand ( x ) ) = x f ( g ( x ) ) = ten and g ( f ( x ) ) = x . g ( f ( ten ) ) = 10 .
  • ⓑ This tells u.s.a. that f f and g 1000 are inverse functions

17 .

f ( g ( x ) ) = x , one thousand ( f ( ten ) ) = ten f ( g ( ten ) ) = x , g ( f ( x ) ) = x

41 .

x x i 4 7 12 16
f − 1 ( x ) f − 1 ( 10 ) 3 six 9 13 14

43 .

f − 1 ( x ) = ( 1 + x ) 1 / 3 f − ane ( 10 ) = ( 1 + 10 ) 1 / 3

Graph of a cubic function and its inverse.

45 .

f − ane ( x ) = 5 ix ( ten − 32 ) . f − 1 ( x ) = 5 9 ( ten − 32 ) . Given the Fahrenheit temperature, x , x , this formula allows yous to calculate the Celsius temperature.

47 .

t ( d ) = d 50 , t ( d ) = d 50 , t ( 180 ) = 180 l . t ( 180 ) = 180 l . The time for the car to travel 180 miles is 3.6 hours.

Review Exercises

5 .

f ( − 3 ) = − 27 ; f ( − 3 ) = − 27 ; f ( two ) = − 2 ; f ( 2 ) = − 2 ; f ( − a ) = − 2 a ii − 3 a ; f ( − a ) = − 2 a 2 − iii a ;
− f ( a ) = 2 a 2 − 3 a ; − f ( a ) = 2 a two − 3 a ; f ( a + h ) = − 2 a 2 + 3 a − 4 a h + 3 h − two h 2 f ( a + h ) = − 2 a 2 + 3 a − iv a h + 3 h − two h 2

17 .

x = − 1.8 ten = − i.8 or  or x = i.8  or 10 = 1.8

xix .

− 64 + lxxx a − 16 a 2 − ane + a = − 16 a + 64 − 64 + 80 a − 16 a two − 1 + a = − 16 a + 64

21 .

( − ∞ , − 2 ) ∪ ( − two , six ) ∪ ( 6 , ∞ ) ( − ∞ , − ii ) ∪ ( − 2 , vi ) ∪ ( 6 , ∞ )

27 .

increasing ( 2 , ∞ ) ; ( two , ∞ ) ; decreasing ( − ∞ , 2 ) ( − ∞ , 2 )

29 .

increasing ( − 3 , 1 ) ; ( − 3 , ane ) ; constant ( − ∞ , − 3 ) ∪ ( 1 , ∞ ) ( − ∞ , − iii ) ∪ ( 1 , ∞ )

31 .

local minimum ( − ii , − 3 ) ; ( − 2 , − 3 ) ; local maximum ( 1 , 3 ) ( 1 , 3 )

33 .

Absolute Maximum: x

35 .

( f ∘ g ) ( x ) = 17 − 18 10 ; ( g ∘ f ) ( 10 ) = − vii − 18 10 ( f ∘ thousand ) ( x ) = 17 − 18 10 ; ( g ∘ f ) ( x ) = − 7 − 18 x

37 .

( f ∘ g ) ( 10 ) = ane 10 + two ; ( f ∘ g ) ( ten ) = 1 x + 2 ; ( g ∘ f ) ( x ) = 1 ten + ii ( thousand ∘ f ) ( x ) = i x + two

39 .

( f ∘ g ) ( x ) = ane + 10 ane + iv x , 10 ≠ 0 , x ≠ − 1 4 ( f ∘ g ) ( x ) = 1 + x ane + 4 x , x ≠ 0 , x ≠ − 1 4

41 .

( f ∘ g ) ( x ) = 1 x , x > 0 ( f ∘ m ) ( x ) = i ten , x > 0

43 .

sample: 1000 ( x ) = 2 x − 1 3 x + 4 ; f ( x ) = x chiliad ( ten ) = 2 x − 1 three x + iv ; f ( x ) = x

55 .

f ( ten ) = | ten − 3 | f ( x ) = | x − 3 |

63 .

f ( x ) = 1 2 | x + two | + 1 f ( x ) = ane 2 | 10 + 2 | + 1

65 .

f ( ten ) = − iii | x − 3 | + 3 f ( x ) = − 3 | ten − 3 | + 3

69 .

x = − 22 , 10 = fourteen x = − 22 , ten = 14

71 .

( − 5 3 , 3 ) ( − 5 iii , iii )

73 .

f − 1 ( x ) = x - i f − 1 ( x ) = x - 1

77 .

The office is one-to-1.

78 .

The function is not one-to-1.

Exercise Test

1 .

The relation is a function.

5 .

The graph is a parabola and the graph fails the horizontal line test.

19 .

x = − vii x = − vii and 10 = x x = 10

21 .

f − 1 ( x ) = ten + 5 3 f − 1 ( x ) = x + five three

23 .

( − ∞ , − 1.ane )  and ( 1.1 , ∞ ) ( − ∞ , − 1.1 )  and ( 1.1 , ∞ )

25 .

( 1.i , − 0.ix ) ( 1.one , − 0.9 )

29 .

f ( x ) = { | ten | if 10 ≤ 2 three if 10 > 2 f ( x ) = { | x | if x ≤ 2 3 if 10 > 2

35 .

f − 1 ( x ) = − 10 − eleven 2 f − 1 ( ten ) = − x − 11 two

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