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5.2E: Exercises for Section 5.2

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In exercises 1 - 4, express the limits as integrals.

1) \(\displaystyle \lim_{n→∞}\sum_{i=1}^n(x^∗_i)Δx\) over \([1,3]\)

2) \(\displaystyle \lim_{n→∞}\sum_{i=1}^n(5(x^∗_i)^2−3(x^∗_i)^3)Δx\) over \([0,2]\)

3) \(\displaystyle \lim_{n→∞}\sum_{i=1}^n\sin^2(2πx^∗_i)Δx\) over \([0,1]\)

4) \(\displaystyle \lim_{n→∞}\sum_{i=1}^n\cos^2(2πx^∗_i)Δx\) over \([0,1]\)

In exercises 5 - 10, given \(L_n\) or \(R_n\) as indicated, express their limits as \(n→∞\) as definite integrals, identifying the correct intervals.

5) \(\displaystyle L_n=\frac{1}{n}\sum_{i=1}^n\frac{i−1}{n}\)

6) \(\displaystyle R_n=\frac{1}{n}\sum_{i=1}^n\frac{i}{n}\)

7) \(\displaystyle Ln=\frac{2}{n}\sum_{i=1}^n(1+2\frac{i−1}{n})\)

8) \(\displaystyle R_n=\frac{3}{n}\sum_{i=1}^n(3+3\frac{i}{n})\)

9) \(\displaystyle L_n=\frac{2π}{n}\sum_{i=1}^n2π\frac{i−1}{n}\cos(2π\frac{i−1}{n})\)

10 \(\displaystyle R_n=\frac{1}{n}\sum_{i=1}^n(1+\frac{i}{n})\log((1+\frac{i}{n})^2)\)

In exercises 11 - 16, evaluate the integrals of the functions graphed using the formulas for areas of triangles and circles, and subtracting the areas below the \(x\) -axis.

In exercises 17 - 24, evaluate the integral using area formulas.

17) \(\displaystyle ∫^3_0(3−x)\,dx\)

18) \(\displaystyle ∫^3_2(3−x)\,dx\)

19) \(\displaystyle ∫^3_{−3}(3−|x|)\,dx\)

20) \(\displaystyle ∫^6_0(3−|x−3|)\,dx\)

21) \(\displaystyle ∫^2_{−2}\sqrt{4−x^2}\,dx\)

22) \(\displaystyle ∫^5_1\sqrt{4−(x−3)^2}\,dx\)

23) \(\displaystyle ∫^{12}_0\sqrt{36−(x−6)^2}\,dx\)

24) \(\displaystyle ∫^3_{−2}(3−|x|)\,dx\)

In exercises 25 - 28, use averages of values at the left (L) and right (R) endpoints to compute the integrals of the piecewise linear functions with graphs that pass through the given list of points over the indicated intervals.

25) \( {(0,0),(2,1),(4,3),(5,0),(6,0),(8,3)}\) over \( [0,8]\)

26) \({(0,2),(1,0),(3,5),(5,5),(6,2),(8,0)}\) over \([0,8]\)

27) \( {(−4,−4),(−2,0),(0,−2),(3,3),(4,3)}\) over \( [−4,4]\)

28) \( {(−4,0),(−2,2),(0,0),(1,2),(3,2),(4,0)}\) over \( [−4,4]\)

Suppose that \(\displaystyle ∫^4_0f(x)\,dx=5\) and \(\displaystyle ∫^2_0f(x)\,dx=−3\) , and \(\displaystyle ∫^4_0g(x)\,dx=−1\) and \(\displaystyle ∫^2_0g(x)\,dx=2\) . In exercises 29 - 34, compute the integrals.

29) \(\displaystyle ∫^4_0(f(x)+g(x))\,dx\)

30) \(\displaystyle ∫^4_2(f(x)+g(x))\,dx\)

31) \(\displaystyle ∫^2_0(f(x)−g(x))\,dx\)

32) \(\displaystyle ∫^4_2(f(x)−g(x))\,dx\)

33) \(\displaystyle ∫^2_0(3f(x)−4g(x))\,dx\)

34) \(\displaystyle ∫^4_2(4f(x)−3g(x))\,dx\)

In exercises 35 - 38, use the identity \(\displaystyle ∫^A_{−A}f(x)\,dx=∫^0_{−A}f(x)\,dx+∫^A_0f(x)\,dx\) to compute the integrals.

35) \(\displaystyle ∫^π_{−π}\frac{\sin t}{1+t^2}dt\) (Hint: \(\displaystyle \sin(−t)=−\sin(t))\)

36) \(\displaystyle ∫^{\sqrt{π}}_\sqrt{−π}\frac{t}{1+\cos t}dt\)

37) \(\displaystyle ∫^3_1(2−x)\,dx\) (Hint: Look at the graph of \(f\).)

38) \(\displaystyle ∫^4_2(x−3)^3\,dx\) (Hint: Look at the graph of \(f\).)

In exercises 39 - 44, given that \(\displaystyle ∫^1_0x\,dx=\frac{1}{2},\;∫^1_0x^2\,dx=\frac{1}{3},\) and \(\displaystyle ∫^1_0x^3\,dx=\frac{1}{4}\) , compute the integrals.

39) \(\displaystyle ∫^1_0(1+x+x^2+x^3)\,dx\)

40) \(\displaystyle ∫^1_0(1−x+x^2−x^3)\,dx\)

41) \(\displaystyle ∫^1_0(1−x)^2\,dx\)

42) \(\displaystyle ∫^1_0(1−2x)^3\,dx\)

43) \(\displaystyle ∫^1_0\left(6x−\tfrac{4}{3}x^2\right)\,dx\)

44) \(\displaystyle ∫^1_0(7−5x^3)\,dx\)

In exercises 45 - 50, use the comparison theorem.

45) Show that \(\displaystyle ∫^3_0(x^2−6x+9)\,dx≥0.\)

46) Show that \(\displaystyle ∫^3_{−2}(x−3)(x+2)\,dx≤0.\)

47) Show that \(\displaystyle ∫^1_0\sqrt{1+x^3}\,dx≤∫^1_0\sqrt{1+x^2}\,dx\).

48) Show that \(\displaystyle ∫^2_1\sqrt{1+x}\,dx≤∫^2_1\sqrt{1+x^2}\,dx.\)

49) Show that \(\displaystyle ∫^{π/2}_0\sin tdt≥\frac{π}{4}\) (Hint: \(\sin t≥\frac{2t}{π}\) over \( [0,\frac{π}{2}])\)

50) Show that \(\displaystyle ∫^{π/4}_{−π/4}\cos t\,dt≥π\sqrt{2}/4\).

In exercises 51 - 56, find the average value \(f_{ave}\) of \(f\) between \(a\) and \(b\) , and find a point \(c\) , where \(f(c)=f_{ave}\)

51) \( f(x)=x^2,\; a=−1,\; b=1\)

52) \( f(x)=x^5,\; a=−1,\; b=1\)

53) \( f(x)=\sqrt{4−x^2},\; a=0,\; b=2\)

54) \(f(x)=3−|x|,\; a=−3,\; b=3\)

55) \(f(x)=\sin x,\; a=0,\; b=2π\)

56) \( f(x)=\cos x,\; a=0,\; b=2π\)

In exercises 57 - 60, approximate the average value using Riemann sums \(L_{100}\) and \(R_{100}\) . How does your answer compare with the exact given answer?

57) [T] \(y=\ln(x)\) over the interval \( [1,4]\); the exact solution is \(\dfrac{\ln(256)}{3}−1.\)

58) [T] \(y=e^{x/2}\) over the interval \([0,1]\); the exact solution is \( 2(\sqrt{e}−1).\)

59) [T] \(y=\tan x\) over the interval \([0,\frac{π}{4}]\); the exact solution is \(\dfrac{2\ln(2)}{π}\).

60) [T] \(y=\dfrac{x+1}{\sqrt{4−x^2}}\) over the interval \([−1,1]\); the exact solution is \(\dfrac{π}{6}\).

In exercises 61 - 64, compute the average value using the left Riemann sums \(L_N\) for \(N=1,10,100\) . How does the accuracy compare with the given exact value?

61) [T] \(y=x^2−4\) over the interval \([0,2]\); the exact solution is \(−\frac{8}{3}\).

62) [T] \(y=xe^{x^2}\) over the interval \([0,2]\); the exact solution is \(\frac{1}{4}(e^4−1).\)

63) [T] \(y=\left(\dfrac{1}{2}\right)^x\) over the interval \([0,4]\); the exact solution is \(\dfrac{15}{64\ln(2)}\).

64) [T] \( y=x\sin(x^2)\) over the interval \( [−π,0]\); the exact solution is \( \dfrac{\cos(π^2)−1}{2π.}\)

65) Suppose that \(\displaystyle A=∫^{2π}_0\sin^2t\,dt\) and \(\displaystyle B=∫^{2π}_0\cos^2t\,dt.\) Show that \(A+B=2π\) and \(A=B.\)

66) Suppose that \(\displaystyle A=∫^{π/4}_{−π/4}\sec^2 t\,dt=π\) and \(\displaystyle B=∫^{π/4}_{−π/}4\tan^2 t\,dt.\) Show that \(A−B=\dfrac{π}{2}\).

67) Show that the average value of \(\sin^2 t\) over \([0,2π]\) is equal to \(1/2.\) Without further calculation, determine whether the average value of \(\sin^2 t\) over \([0,π]\) is also equal to \(1/2.\)

68) Show that the average value of \(\cos^2 t\) over \([0,2π]\) is equal to \(1/2.\) Without further calculation, determine whether the average value of \(\cos^2(t)\) over \([0,π]\) is also equal to \(1/2.\)

69) Explain why the graphs of a quadratic function (parabola) \(p(x)\) and a linear function \(ℓ(x)\) can intersect in at most two points. Suppose that \(p(a)=ℓ(a)\) and \(p(b)=ℓ(b)\), and that \(\displaystyle ∫^b_ap(t)\,dt>∫^b_aℓ(t)dt\). Explain why \(\displaystyle ∫^d_cp(t)>∫^d_cℓ(t)\,dt\) whenever \( a≤c<d≤b.\)

70) Suppose that parabola \(p(x)=ax^2+bx+c\) opens downward \((a<0)\) and has a vertex of \(y=\dfrac{−b}{2a}>0\). For which interval \([A,B]\) is \(\displaystyle ∫^B_A(ax^2+bx+c)\,dx\) as large as possible?

71) Suppose \([a,b]\) can be subdivided into subintervals \(a=a_0<a_1<a_2<⋯<a_N=b\) such that either \(f≥0\) over \([a_{i−1},a_i]\) or \(f≤0\) over \([a_{i−1},a_i]\). Set \(\displaystyle A_i=∫^{a_i}_{a_{i−1}}f(t)\,dt.\)

a. Explain why \(\displaystyle ∫^b_af(t)\,dt=A_1+A_2+⋯+A_N.\)

b. Then, explain why \(\displaystyle ∫^b_af(t)\,dt≤∫^b_a|f(t)|\,dt.\)

72) Suppose \(f\) and \(g\) are continuous functions such that \(\displaystyle ∫^d_cf(t)\,dt≤∫^d_cg(t)\,dt\) for every subinterval \([c,d]\) of \([a,b]\). Explain why \( f(x)≤g(x)\) for all values of \(x.\)

73) Suppose the average value of \(f\) over \([a,b]\) is \(1\) and the average value of \(f\) over \([b,c]\) is \(1\) where \(a<c<b\). Show that the average value of \(f\) over \([a,c]\) is also \(1.\)

74) Suppose that \([a,b]\) can be partitioned. taking \(a=a_0<a_1<⋯<a_N=b\) such that the average value of \(f\) over each subinterval \([a_{i−1},a_i]=1\) is equal to 1 for each \( i=1,…,N\). Explain why the average value of f over \( [a,b]\) is also equal to \(1.\)

75) Suppose that for each \(i\) such that \( 1≤i≤N\) one has \(\displaystyle ∫^i_{i−1}f(t)\,dt=i\). Show that \(\displaystyle ∫^N_0f(t)\,dt=\frac{N(N+1)}{2}.\)

76) Suppose that for each \(i\) such that \(1≤i≤N\) one has \(\displaystyle ∫^i_{i−1}f(t)\,dt=i^2\). Show that \(\displaystyle ∫^N_0f(t)\,dt=\frac{N(N+1)(2N+1)}{6}\).

77) [T] Compute the left and right Riemann sums \(\displaystyle L_{10}\) and \(R_{10}\) and their average \(\dfrac{L_{10}+R_{10}}{2}\) for \( f(t)=t^2\)over \( [0,1]\). Given that \(\displaystyle ∫^1_0t^2\,dt=1/3\), to how many decimal places is \( \dfrac{L_{10}+R_{10}}{2}\) accurate?

78) [T] Compute the left and right Riemann sums, \(L_10\) and \(R_{10}\), and their average \(\dfrac{L_{10}+R_{10}}{2}\) for \( f(t)=(4−t^2)\) over \([1,2]\). Given that \(\displaystyle ∫^2_1(4−t^2)\,dt=1.66\), to how many decimal places is \(\dfrac{L_{10}+R_{10}}{2}\) accurate?

79) If \(\displaystyle ∫^5_1\sqrt{1+t^4}\,dt=41.7133...,\) what is \(\displaystyle ∫^5_1\sqrt{1+u^4}\,du?\)

80) Estimate \(\displaystyle ∫^1_0t\,dt\) using the left and right endpoint sums, each with a single rectangle. How does the average of these left and right endpoint sums compare with the actual value \(\displaystyle ∫^1_0t\,dt?\)

81) Estimate \(\displaystyle ∫^1_0t\,dt\) by comparison with the area of a single rectangle with height equal to the value of \(t\) at the midpoint \(t=\dfrac{1}{2}\). How does this midpoint estimate compare with the actual value \(\displaystyle ∫^1_0t\,dt?\)

82) From the graph of \(\sin(2πx)\) shown:

a. Explain why \(\displaystyle ∫^1_0\sin(2πt)\,dt=0.\)

b. Explain why, in general, \(\displaystyle ∫^{a+1}_a\sin(2πt)\,dt=0\) for any value of \(a\).

a. The graph is antisymmetric with respect to \(t=\frac{1}{2}\) over \([0,1]\), so the average value is zero. b. For any value of \(a\), the graph between \([a,a+1]\) is a shift of the graph over \([0,1]\), so the net areas above and below the axis do not change and the average remains zero.

83) If f is 1-periodic \((f(t+1)=f(t))\), odd, and integrable over \([0,1]\), is it always true that \(\displaystyle ∫^1_0f(t)\,dt=0?\)

84) If f is 1-periodic and \(\displaystyle ∫10f(t)\,dt=A,\) is it necessarily true that \(\displaystyle ∫^{1+a}_af(t)\,dt=A\) for all \(A\)?

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2.1 The Rectangular Coordinate Systems and Graphs

x -intercept is ( 4 , 0 ) ; ( 4 , 0 ) ; y- intercept is ( 0 , 3 ) . ( 0 , 3 ) .

125 = 5 5 125 = 5 5

( − 5 , 5 2 ) ( − 5 , 5 2 )

2.2 Linear Equations in One Variable

x = −5 x = −5

x = −3 x = −3

x = 10 3 x = 10 3

x = 1 x = 1

x = − 7 17 . x = − 7 17 . Excluded values are x = − 1 2 x = − 1 2 and x = − 1 3 . x = − 1 3 .

x = 1 3 x = 1 3

m = − 2 3 m = − 2 3

y = 4 x −3 y = 4 x −3

x + 3 y = 2 x + 3 y = 2

Horizontal line: y = 2 y = 2

Parallel lines: equations are written in slope-intercept form.

y = 5 x + 3 y = 5 x + 3

2.3 Models and Applications

C = 2.5 x + 3 , 650 C = 2.5 x + 3 , 650

L = 37 L = 37 cm, W = 18 W = 18 cm

2.4 Complex Numbers

−24 = 0 + 2 i 6 −24 = 0 + 2 i 6

( 3 −4 i ) − ( 2 + 5 i ) = 1 −9 i ( 3 −4 i ) − ( 2 + 5 i ) = 1 −9 i

5 2 − i 5 2 − i

18 + i 18 + i

−3 −4 i −3 −4 i

2.5 Quadratic Equations

( x − 6 ) ( x + 1 ) = 0 ; x = 6 , x = − 1 ( x − 6 ) ( x + 1 ) = 0 ; x = 6 , x = − 1

( x −7 ) ( x + 3 ) = 0 , ( x −7 ) ( x + 3 ) = 0 , x = 7 , x = 7 , x = −3. x = −3.

( x + 5 ) ( x −5 ) = 0 , ( x + 5 ) ( x −5 ) = 0 , x = −5 , x = −5 , x = 5. x = 5.

( 3 x + 2 ) ( 4 x + 1 ) = 0 , ( 3 x + 2 ) ( 4 x + 1 ) = 0 , x = − 2 3 , x = − 2 3 , x = − 1 4 x = − 1 4

x = 0 , x = −10 , x = −1 x = 0 , x = −10 , x = −1

x = 4 ± 5 x = 4 ± 5

x = 3 ± 22 x = 3 ± 22

x = − 2 3 , x = − 2 3 , x = 1 3 x = 1 3

2.6 Other Types of Equations

{ −1 } { −1 }

0 , 0 , 1 2 , 1 2 , − 1 2 − 1 2

1 ; 1 ; extraneous solution − 2 9 − 2 9

−2 ; −2 ; extraneous solution −1 −1

−1 , −1 , 3 2 3 2

−3 , 3 , − i , i −3 , 3 , − i , i

2 , 12 2 , 12

−1 , −1 , 0 0 is not a solution.

2.7 Linear Inequalities and Absolute Value Inequalities

[ −3 , 5 ] [ −3 , 5 ]

( − ∞ , −2 ) ∪ [ 3 , ∞ ) ( − ∞ , −2 ) ∪ [ 3 , ∞ )

x < 1 x < 1

x ≥ −5 x ≥ −5

( 2 , ∞ ) ( 2 , ∞ )

[ − 3 14 , ∞ ) [ − 3 14 , ∞ )

6 < x ≤ 9 ​ or ( 6 , 9 ] 6 < x ≤ 9 ​ or ( 6 , 9 ]

( − 1 8 , 1 2 ) ( − 1 8 , 1 2 )

| x −2 | ≤ 3 | x −2 | ≤ 3

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

2.1 Section Exercises

Answers may vary. Yes. It is possible for a point to be on the x -axis or on the y -axis and therefore is considered to NOT be in one of the quadrants.

The y -intercept is the point where the graph crosses the y -axis.

The x- intercept is ( 2 , 0 ) ( 2 , 0 ) and the y -intercept is ( 0 , 6 ) . ( 0 , 6 ) .

The x- intercept is ( 2 , 0 ) ( 2 , 0 ) and the y -intercept is ( 0 , −3 ) . ( 0 , −3 ) .

The x- intercept is ( 3 , 0 ) ( 3 , 0 ) and the y -intercept is ( 0 , 9 8 ) . ( 0 , 9 8 ) .

y = 4 − 2 x y = 4 − 2 x

y = 5 − 2 x 3 y = 5 − 2 x 3

y = 2 x − 4 5 y = 2 x − 4 5

d = 74 d = 74

d = 36 = 6 d = 36 = 6

d ≈ 62.97 d ≈ 62.97

( 3 , − 3 2 ) ( 3 , − 3 2 )

( 2 , −1 ) ( 2 , −1 )

( 0 , 0 ) ( 0 , 0 )

y = 0 y = 0

not collinear

A: ( −3 , 2 ) , B: ( 1 , 3 ) , C: ( 4 , 0 ) A: ( −3 , 2 ) , B: ( 1 , 3 ) , C: ( 4 , 0 )

d = 8.246 d = 8.246

d = 5 d = 5

( −3 , 4 ) ( −3 , 4 )

x = 0          y = −2 x = 0          y = −2

x = 0.75 y = 0 x = 0.75 y = 0

x = − 1.667 y = 0 x = − 1.667 y = 0

15 − 11.2 = 3.8 mi 15 − 11.2 = 3.8 mi shorter

6 .0 42 6 .0 42

Midpoint of each diagonal is the same point ( 2 , –2 ) ( 2 , –2 ) . Note this is a characteristic of rectangles, but not other quadrilaterals.

2.2 Section Exercises

It means they have the same slope.

The exponent of the x x variable is 1. It is called a first-degree equation.

If we insert either value into the equation, they make an expression in the equation undefined (zero in the denominator).

x = 2 x = 2

x = 2 7 x = 2 7

x = 6 x = 6

x = 3 x = 3

x = −14 x = −14

x ≠ −4 ; x ≠ −4 ; x = −3 x = −3

x ≠ 1 ; x ≠ 1 ; when we solve this we get x = 1 , x = 1 , which is excluded, therefore NO solution

x ≠ 0 ; x ≠ 0 ; x = − 5 2 x = − 5 2

y = − 4 5 x + 14 5 y = − 4 5 x + 14 5

y = − 3 4 x + 2 y = − 3 4 x + 2

y = 1 2 x + 5 2 y = 1 2 x + 5 2

y = −3 x − 5 y = −3 x − 5

y = 7 y = 7

y = −4 y = −4

8 x + 5 y = 7 8 x + 5 y = 7

Perpendicular

m = − 9 7 m = − 9 7

m = 3 2 m = 3 2

m 1 = − 1 3 ,   m 2 = 3 ;   Perpendicular . m 1 = − 1 3 ,   m 2 = 3 ;   Perpendicular .

y = 0.245 x − 45.662. y = 0.245 x − 45.662. Answers may vary. y min = −50 , y max = −40 y min = −50 , y max = −40

y = − 2.333 x + 6.667. y = − 2.333 x + 6.667. Answers may vary. y min = −10 ,   y max = 10 y min = −10 ,   y max = 10

y = − A B x + C B y = − A B x + C B

The slope for  ( −1 , 1 ) to  ( 0 , 4 ) is  3. The slope for  ( −1 , 1 ) to  ( 2 , 0 ) is  − 1 3 . The slope for  ( 2 , 0 ) to  ( 3 , 3 ) is  3. The slope for  ( 0 , 4 ) to  ( 3 , 3 ) is  − 1 3 . The slope for  ( −1 , 1 ) to  ( 0 , 4 ) is  3. The slope for  ( −1 , 1 ) to  ( 2 , 0 ) is  − 1 3 . The slope for  ( 2 , 0 ) to  ( 3 , 3 ) is  3. The slope for  ( 0 , 4 ) to  ( 3 , 3 ) is  − 1 3 .

Yes they are perpendicular.

2.3 Section Exercises

Answers may vary. Possible answers: We should define in words what our variable is representing. We should declare the variable. A heading.

2 , 000 − x 2 , 000 − x

v + 10 v + 10

Ann: 23 ; 23 ; Beth: 46 46

20 + 0.05 m 20 + 0.05 m

90 + 40 P 90 + 40 P

50 , 000 − x 50 , 000 − x

She traveled for 2 h at 20 mi/h, or 40 miles.

$5,000 at 8% and $15,000 at 12%

B = 100 + .05 x B = 100 + .05 x

R = 9 R = 9

r = 4 5 r = 4 5 or 0.8

W = P − 2 L 2 = 58 − 2 ( 15 ) 2 = 14 W = P − 2 L 2 = 58 − 2 ( 15 ) 2 = 14

f = p q p + q = 8 ( 13 ) 8 + 13 = 104 21 f = p q p + q = 8 ( 13 ) 8 + 13 = 104 21

m = − 5 4 m = − 5 4

h = 2 A b 1 + b 2 h = 2 A b 1 + b 2

length = 360 ft; width = 160 ft

A = 88 in . 2 A = 88 in . 2

h = V π r 2 h = V π r 2

r = V π h r = V π h

C = 12 π C = 12 π

2.4 Section Exercises

Add the real parts together and the imaginary parts together.

Possible answer: i i times i i equals -1, which is not imaginary.

−8 + 2 i −8 + 2 i

14 + 7 i 14 + 7 i

− 23 29 + 15 29 i − 23 29 + 15 29 i

8 − i 8 − i

−11 + 4 i −11 + 4 i

2 −5 i 2 −5 i

6 + 15 i 6 + 15 i

−16 + 32 i −16 + 32 i

−4 −7 i −4 −7 i

2 − 2 3 i 2 − 2 3 i

4 − 6 i 4 − 6 i

2 5 + 11 5 i 2 5 + 11 5 i

1 + i 3 1 + i 3

( 3 2 + 1 2 i ) 6 = −1 ( 3 2 + 1 2 i ) 6 = −1

5 −5 i 5 −5 i

9 2 − 9 2 i 9 2 − 9 2 i

2.5 Section Exercises

It is a second-degree equation (the highest variable exponent is 2).

We want to take advantage of the zero property of multiplication in the fact that if a ⋅ b = 0 a ⋅ b = 0 then it must follow that each factor separately offers a solution to the product being zero: a = 0 o r b = 0. a = 0 o r b = 0.

One, when no linear term is present (no x term), such as x 2 = 16. x 2 = 16. Two, when the equation is already in the form ( a x + b ) 2 = d . ( a x + b ) 2 = d .

x = 6 , x = 6 , x = 3 x = 3

x = − 5 2 , x = − 5 2 , x = − 1 3 x = − 1 3

x = 5 , x = 5 , x = −5 x = −5

x = − 3 2 , x = − 3 2 , x = 3 2 x = 3 2

x = −2 , 3 x = −2 , 3

x = 0 , x = 0 , x = − 3 7 x = − 3 7

x = −6 , x = −6 , x = 6 x = 6

x = 6 , x = 6 , x = −4 x = −4

x = 1 , x = 1 , x = −2 x = −2

x = −2 , x = −2 , x = 11 x = 11

z = 2 3 , z = 2 3 , z = − 1 2 z = − 1 2

x = 3 ± 17 4 x = 3 ± 17 4

One rational

Two real; rational

x = − 1 ± 17 2 x = − 1 ± 17 2

x = 5 ± 13 6 x = 5 ± 13 6

x = − 1 ± 17 8 x = − 1 ± 17 8

x ≈ 0.131 x ≈ 0.131 and x ≈ 2.535 x ≈ 2.535

x ≈ − 6.7 x ≈ − 6.7 and x ≈ 1.7 x ≈ 1.7

a x 2 + b x + c = 0 x 2 + b a x = − c a x 2 + b a x + b 2 4 a 2 = − c a + b 4 a 2 ( x + b 2 a ) 2 = b 2 − 4 a c 4 a 2 x + b 2 a = ± b 2 − 4 a c 4 a 2 x = − b ± b 2 − 4 a c 2 a a x 2 + b x + c = 0 x 2 + b a x = − c a x 2 + b a x + b 2 4 a 2 = − c a + b 4 a 2 ( x + b 2 a ) 2 = b 2 − 4 a c 4 a 2 x + b 2 a = ± b 2 − 4 a c 4 a 2 x = − b ± b 2 − 4 a c 2 a

x ( x + 10 ) = 119 ; x ( x + 10 ) = 119 ; 7 ft. and 17 ft.

maximum at x = 70 x = 70

The quadratic equation would be ( 100 x −0.5 x 2 ) − ( 60 x + 300 ) = 300. ( 100 x −0.5 x 2 ) − ( 60 x + 300 ) = 300. The two values of x x are 20 and 60.

2.6 Section Exercises

This is not a solution to the radical equation, it is a value obtained from squaring both sides and thus changing the signs of an equation which has caused it not to be a solution in the original equation.

He or she is probably trying to enter negative 9, but taking the square root of −9 −9 is not a real number. The negative sign is in front of this, so your friend should be taking the square root of 9, cubing it, and then putting the negative sign in front, resulting in −27. −27.

A rational exponent is a fraction: the denominator of the fraction is the root or index number and the numerator is the power to which it is raised.

x = 81 x = 81

x = 17 x = 17

x = 8 ,     x = 27 x = 8 ,     x = 27

x = −2 , 1 , −1 x = −2 , 1 , −1

y = 0 ,     3 2 ,     − 3 2 y = 0 ,     3 2 ,     − 3 2

m = 1 , −1 m = 1 , −1

x = 2 5 , ±3 i x = 2 5 , ±3 i

x = 32 x = 32

t = 44 3 t = 44 3

x = −2 x = −2

x = 4 , −4 3 x = 4 , −4 3

x = − 5 4 , 7 4 x = − 5 4 , 7 4

x = 3 , −2 x = 3 , −2

x = 1 , −1 , 3 , -3 x = 1 , −1 , 3 , -3

x = 2 , −2 x = 2 , −2

x = 1 , 5 x = 1 , 5

x ≥ 0 x ≥ 0

x = 4 , 6 , −6 , −8 x = 4 , 6 , −6 , −8

2.7 Section Exercises

When we divide both sides by a negative it changes the sign of both sides so the sense of the inequality sign changes.

( − ∞ , ∞ ) ( − ∞ , ∞ )

We start by finding the x -intercept, or where the function = 0. Once we have that point, which is ( 3 , 0 ) , ( 3 , 0 ) , we graph to the right the straight line graph y = x −3 , y = x −3 , and then when we draw it to the left we plot positive y values, taking the absolute value of them.

( − ∞ , 3 4 ] ( − ∞ , 3 4 ]

[ − 13 2 , ∞ ) [ − 13 2 , ∞ )

( − ∞ , 3 ) ( − ∞ , 3 )

( − ∞ , − 37 3 ] ( − ∞ , − 37 3 ]

All real numbers ( − ∞ , ∞ ) ( − ∞ , ∞ )

( − ∞ , − 10 3 ) ∪ ( 4 , ∞ ) ( − ∞ , − 10 3 ) ∪ ( 4 , ∞ )

( − ∞ , −4 ] ∪ [ 8 , + ∞ ) ( − ∞ , −4 ] ∪ [ 8 , + ∞ )

No solution

( −5 , 11 ) ( −5 , 11 )

[ 6 , 12 ] [ 6 , 12 ]

[ −10 , 12 ] [ −10 , 12 ]

x > − 6 and x > − 2 Take the intersection of two sets . x > − 2 ,   ( − 2 , + ∞ ) x > − 6 and x > − 2 Take the intersection of two sets . x > − 2 ,   ( − 2 , + ∞ )

x < − 3   or   x ≥ 1 Take the union of the two sets . ( − ∞ , − 3 ) ∪ ​ ​ [ 1 , ∞ ) x < − 3   or   x ≥ 1 Take the union of the two sets . ( − ∞ , − 3 ) ∪ ​ ​ [ 1 , ∞ )

( − ∞ , −1 ) ∪ ( 3 , ∞ ) ( − ∞ , −1 ) ∪ ( 3 , ∞ )

[ −11 , −3 ] [ −11 , −3 ]

It is never less than zero. No solution.

Where the blue line is above the orange line; point of intersection is x = − 3. x = − 3.

( − ∞ , −3 ) ( − ∞ , −3 )

Where the blue line is above the orange line; always. All real numbers.

( − ∞ , − ∞ ) ( − ∞ , − ∞ )

( −1 , 3 ) ( −1 , 3 )

( − ∞ , 4 ) ( − ∞ , 4 )

{ x | x < 6 } { x | x < 6 }

{ x | −3 ≤ x < 5 } { x | −3 ≤ x < 5 }

( −2 , 1 ] ( −2 , 1 ]

( − ∞ , 4 ] ( − ∞ , 4 ]

Where the blue is below the orange; always. All real numbers. ( − ∞ , + ∞ ) . ( − ∞ , + ∞ ) .

Where the blue is below the orange; ( 1 , 7 ) . ( 1 , 7 ) .

x = 2 , − 4 5 x = 2 , − 4 5

( −7 , 5 ] ( −7 , 5 ]

80 ≤ T ≤ 120 1 , 600 ≤ 20 T ≤ 2 , 400 80 ≤ T ≤ 120 1 , 600 ≤ 20 T ≤ 2 , 400

[ 1 , 600 , 2 , 400 ] [ 1 , 600 , 2 , 400 ]

Review Exercises

x -intercept: ( 3 , 0 ) ; ( 3 , 0 ) ; y -intercept: ( 0 , −4 ) ( 0 , −4 )

y = 5 3 x + 4 y = 5 3 x + 4

72 = 6 2 72 = 6 2

620.097 620.097

midpoint is ( 2 , 23 2 ) ( 2 , 23 2 )

x = 4 x = 4

x = 12 7 x = 12 7

y = 1 6 x + 4 3 y = 1 6 x + 4 3

y = 2 3 x + 6 y = 2 3 x + 6

females 17, males 56

x = − 3 4 ± i 47 4 x = − 3 4 ± i 47 4

horizontal component −2 ; −2 ; vertical component −1 −1

7 + 11 i 7 + 11 i

−16 − 30 i −16 − 30 i

−4 − i 10 −4 − i 10

x = 7 − 3 i x = 7 − 3 i

x = −1 , −5 x = −1 , −5

x = 0 , 9 7 x = 0 , 9 7

x = 10 , −2 x = 10 , −2

x = − 1 ± 5 4 x = − 1 ± 5 4

x = 2 5 , − 1 3 x = 2 5 , − 1 3

x = 5 ± 2 7 x = 5 ± 2 7

x = 0 , 256 x = 0 , 256

x = 0 , ± 2 x = 0 , ± 2

x = 11 2 , −17 2 x = 11 2 , −17 2

[ − 10 3 , 2 ] [ − 10 3 , 2 ]

( − 4 3 , 1 5 ) ( − 4 3 , 1 5 )

Where the blue is below the orange line; point of intersection is x = 3.5. x = 3.5.

( 3.5 , ∞ ) ( 3.5 , ∞ )

Practice Test

y = 3 2 x + 2 y = 3 2 x + 2

( 0 , −3 ) ( 0 , −3 ) ( 4 , 0 ) ( 4 , 0 )

( − ∞ , 9 ] ( − ∞ , 9 ]

x = −15 x = −15

x ≠ −4 , 2 ; x ≠ −4 , 2 ; x = − 5 2 , 1 x = − 5 2 , 1

x = 3 ± 3 2 x = 3 ± 3 2

( −4 , 1 ) ( −4 , 1 )

y = −5 9 x − 2 9 y = −5 9 x − 2 9

y = 5 2 x − 4 y = 5 2 x − 4

5 13 − 14 13 i 5 13 − 14 13 i

x = 2 , − 4 3 x = 2 , − 4 3

x = 1 2 ± 2 2 x = 1 2 ± 2 2

x = 1 2 , 2 , −2 x = 1 2 , 2 , −2

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Access for free at https://openstax.org/books/college-algebra/pages/1-introduction-to-prerequisites

• Authors: Jay Abramson
• Publisher/website: OpenStax
• Book title: College Algebra
• Publication date: Feb 13, 2015
• Location: Houston, Texas
• Book URL: https://openstax.org/books/college-algebra/pages/1-introduction-to-prerequisites
• Section URL: https://openstax.org/books/college-algebra/pages/chapter-2

© Dec 8, 2021 OpenStax. Textbook content produced by OpenStax is licensed under a Creative Commons Attribution License . The OpenStax name, OpenStax logo, OpenStax book covers, OpenStax CNX name, and OpenStax CNX logo are not subject to the Creative Commons license and may not be reproduced without the prior and express written consent of Rice University.

Math 6290 (Spring 2020): Homological algebra

The textbook will be An introduction to homological algebra , by Charles A. Weibel.

Lecture recordings

• April 24 : Galois cohomology
• April 14 : group cohomology
• April 10 : derived categories
• April 6 : derived functors
• April 3 : triangulated categories
• For Wednesday, April 29: We will discuss homotopy limits and colimits and stable infinity categories, summarizing HTT , §§1.2.8, 1.2.12, 1.2.13 and HA , §§1.1.1, 1.1.2, 1.3.1, 1.3.2.
• For Wednesday, April 22: Do Exercises 6.2.2 , 6.2.4, 6.4.1, 6.4.2 , 6.6.2 , 6.6.6 .
• For Monday, April 18: Read Weibel, §§6.5, 6.6.
• For Wednesday, April 15: Read Weibel, §§6.2, 6.4. Do Exercises 6.1.6 , 6.2.3 .
• For Monday, April 13: Read Weibel, §6.1. For Friday, April 17: Do Weibel, Exercises 10.7.1 , 10.7.2 , 6.1.1, 6.1.2 , 6.1.3 .
• For Friday, April 10: Read Weibel, §§10.7, 10.8.
• For Wednesday, April 8: Do Weibel, Exercise 10.4.2 .
• Friday, April 3: class will meet on Zoom in Meeting Room 283-728-921 . For Wednesday, April 8: Do Weibel, Exercises 10.1.2 , 10.2.1 , 10.2.2 , 10.2.4, 10.2.6, 10.4.5, 10.4.6.
• Wednesday, April 1: class will meet on Zoom in Meeting Room 283-728-921 . Read Weibel, §10.2. Do Weibel, Exercise 10.3.1 (using this outline ).
• Monday, March 30: class will meet on Zoom in Meeting Room 283-728-921 . We will continue discussing the derived category.
• Friday, March 20: class will meet on Zoom in Meeting Room 283-728-921 . Read Weibel, §10.3 and §10.4. Do Exercise 10.4.2, 10.4.5.
• Wednesday, March 18: class will meet on Zoom in Meeting Room 283-728-921 . We will discuss Exercises 5.6.2, 1.4.5, and the exercise from class on Monday: show that if \( \varphi : A^\bullet \to B^\bullet \) is a quasi-isomorphism of bounded below cochain complexes and \( \alpha : A^\bullet \to I^\bullet \) is a morphism, where \( I^\bullet \) is a bounded below cochain complex of injectives, then there is a morphism of complexes \( \beta : B^\bullet \to I^\bullet \) such that \( \beta \varphi \) is homotopy equivalent to \( \alpha \).
• Friday, March 13: class will meet on Zoom in Meeting Room 283-728-921 . Please read Weibel, §5.8 (skip 5.8.6 unless you already know about sheaves; refer back to §5.7 for the definition of a Cartan–Eilenberg resolution, but you don't need to read the whole section). Do Exercise 5.6.2.
• Wednesday, March 4: (for Friday, March 6) read Weibel, §§5.2, 5.4. (for Wednesday, March 11) do Weibel, Exercises 3.5.1 , 5.2.3 , 5.4.2, 5.4.4 . Do Weibel, Exercises 3.1.1 again, this time using a spectral sequence.
• Wednesday, February 26: (for Wednesday, March 4) do Weibel, Exercises 3.5.1 , 5.2.1 , 5.2.2 .
• Monday, February 17: (for Friday, February 21) read Weibel, §§2.6, 3.5. (to discuss on Wednesday, February 26) Do Weibel, Exercises 2.6.4, 2.6.5, 3.5.1 , 3.5.2 , 3.5.5 .
• Friday, February 14: (for Monday, February 17) read Weibel, §3.4. (to discuss on Wednesday, February 19) Prove that \( R^n \operatorname{Hom}(N,-)(M) \) is a universal \( \delta \)-functor of the variable \( N \) and conclude that \( R^n \operatorname{Hom}(N,-)(M) = R^n \operatorname{Hom}(-,M)(N) \) for all \( n \). In class we defined \( \underline{\operatorname{Ext}}^1(M,N) \) to be the set of isomorphism classes of extensions of \( M \) by \( N \). Define \( \underline{\operatorname{Ext}}^0(M,N) = \operatorname{Hom}(M,N) \) and show that \( \underline{\operatorname{Ext}}^\bullet(M,N) \) is an effaceable \( \delta \)-functor in degrees \( \leq 1 \). Conclude that \( \underline{\operatorname{Ext}}^1(M,N) = \operatorname{Ext}^1(M,N) \).
• Wednesday, February 5: (for Friday, February 7) read Weibel, §2.6, up to page 53 (stop before Application 2.6.5).
• Monday, February 3: (for Friday, February 7) read Weibel, §§2.1, 2.4, 2.5. (to discuss on Wednesday, February 12) Weibel, Exercises 2.1.2, 2.4.2, 2.4.3 , 2.4.5.
• Wednesday, January 29: (to discuss on Wednesday, February 5) Weibel, Exercises 2.2.1, 2.3.2 .
• Monday, January 27: (for Friday, January 31) Read Weibel, §2.2–2.3 (to discuss on Wednesday, February 5) Weibel, Exercises 1.5.3, 1.5.6, 1.5.8 .
• Friday, January 24 (problems to be discussed on Wednesday, January 29): Read Weibel, §1.5. Let \( \operatorname{Hom}_\bullet(A_\bullet, B_\bullet) \) be defined as in class. Show that it is a complex and that \( Z_0 \operatorname{Hom}_\bullet(A_\bullet, B_\bullet) = \operatorname{Hom}(A_\bullet, B_\bullet) \). Do exercises 1.4.1 , 1.4.2, 1.4.3, 1.4.4 , 1.4.5 .
• Wednesday, January 22 (problems to be discussed on Wednesday, January 29): Read Weibel, §1.4. Do exercises 1.2.5, 1.3.1 , 1.3.2, 1.3.5 .
• Friday, January 17 (to be discussed on Wednesday, January 22): Read Weibel, §1.3. Do Weibel, Ex. 1.2.7, 1.2.8 , 1.3.1. Please let me know before class if you do not want to participate in the presentations in class on Wednesday.
• Wednesday, January 15 (to be discussed on Wednesday, January 22): Read Weibel, §1.1 – 1.2. Complete the construction of the abelian group structure on \( \operatorname{Hom}_{\scr A}(X,Y) \) when \( X \) and \( Y \) are objects of an abelian category \( \scr A \) : verify that the addition law is associative and commutative with \( 0 \) as identity and that composition is bi-additive. Do Weibel, Ex. 1.1.1, 1.1.5, 1.1.6 , 1.1.7.
• Monday, January 13 (to be discussed on Wednesday, January 22): Read §A.4 through Example A.4.4 (pp. 424 — 426). Verify that, for any not-necessarily-commutative ring \( R \), the category of left \( R \)-modules is an abelian category. You may use the definition of an abelian category given in Weibel or the definition given in class (but please be clear about which one you are using!).

Office hours

To get an A in this course, you should participate actively in class, and present homework solutions when called upon to do so (or submit written homework solutions).