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Mathematics LibreTexts

3.1E: Exercises

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Exercises

In Exercises 1 - 10, find the degree, the leading term, the leading coefficient, the constant term and the end behavior of the given polynomial.

  1. f(x)=4x3x2
  2. g(x)=3x52x2+x+1
  3. q(r)=116r4
  4. Z(b)=42bb3
  5. f(x)=3x17+22.5x10πx7+13
  6. s(t)=4.9t2+v0t+s0
  7. P(x)=(x1)(x2)(x3)(x4)
  8. p(t)=t2(35t)(t2+t+4)
  9. f(x)=2x3(x+1)(x+2)2
  10. G(t)=4(t2)2(t+12)

In Exercises 11 - 20, find the real zeros of the given polynomial and their corresponding multiplicities. Use this information along with a sign chart to provide a rough sketch of the graph of the polynomial. Compare your answer with the result from a graphing utility.

  1. a(x)=x(x+2)2
  2. g(x)=x(x+2)3
  3. f(x)=2(x2)2(x+1)
  4. g(x)=(2x+1)2(x3)
  5. F(x)=x3(x+2)2
  6. P(x)=(x1)(x2)(x3)(x4)
  7. Q(x)=(x+5)2(x3)4
  8. h(x)=x2(x2)2(x+2)2
  9. H(t)=(3t)(t2+1)
  10. Z(b)=b(42b2)

In Exercises 21 - 26, given the pair of functions f and g, sketch the graph of y=g(x) by starting with the graph of y=f(x) and using transformations. Track at least three points of your choice through the transformations. State the domain and range of g.

  1. f(x)=x3, g(x)=(x+2)3+1
  2. f(x)=x4, g(x)=(x+2)4+1
  3. f(x)=x4, g(x)=23(x1)4
  4. f(x)=x5, g(x)=x53
  5. f(x)=x5, g(x)=(x+1)5+10
  6. f(x)=x6, g(x)=8x6
  7. Use the Intermediate Value Theorem to prove that f(x)=x39x+5 has a real zero in each of the following intervals: [4,3],[0,1] and [2,3].
  8. Rework Example 3.1.3 assuming the box is to be made from an 8.5 inch by 11 inch sheet of paper. Using scissors and tape, construct the box. Are you surprised?

In Exercises 29 - 31, suppose the revenue R, in thousands of dollars, from producing and selling x hundred LCD TVs is given by R(x)=5x3+35x2+155x for 0x10.07.

  1. Use a graphing utility to graph y=R(x) and determine the number of TVs which should be sold to maximize revenue. What is the maximum revenue?
  2. Assume that the cost, in thousands of dollars, to produce x hundred LCD TVs is given by C(x)=200x+25 for x0. Find and simplify an expression for the profit function P(x). (Remember: Profit = Revenue - Cost.)
  3. Use a graphing utility to graph y=P(x) and determine the number of TVs which should be sold to maximize profit. What is the maximum profit?
  4. While developing their newest game, Sasquatch Attack!, the makers of the PortaBoy (from Example 2.1.5) revised their cost function and now use C(n)=.03n34.5n2+225n+250, for n0. As before, C(n) is the cost to make n PortaBoy Game Systems. Market research indicates that the demand function p(x)=1.5n+250 remains unchanged. Use graphing technology to find the production level n that maximizes the profit made by producing and selling n PortaBoy game systems.
  5. According to US Postal regulations, a rectangular shipping box must satisfy the inequality “Length + Girth 130 inches” for Parcel Post and “Length + Girth 108 inches” for other services. Let’s assume we have a closed rectangular box with a square face of side length x as drawn below. The length is the longest side and is clearly labeled. The girth is the distance around the box in the other two dimensions so in our case it is the sum of the four sides of the square, 4x.
    1. Assuming that we’ll be mailing a box via Parcel Post where Length + Girth = 130 inches, express the length of the box in terms of x and then express the volume V of the box in terms of x.
    2. Find the dimensions of the box of maximum volume that can be shipped via Parcel Post.
    3. Repeat parts 33a and 33b if the box is shipped using “other services”.

      Screen Shot 2022-03-28 at 4.00.40 PM.png

  6. [OMITTED]
  7. [OMITTED]
  8. Show that the end behavior of a linear function f(x)=mx+b is as it should be according to the results we’ve established in the section for polynomials of odd degree. (That is, show that the graph of a linear function is “up on one side and down on the other” just like the graph of y=anxn for odd numbers n.)
  9. There is one subtlety about the role of multiplicity that we need to discuss further; specifically we need to see ‘how’ the graph crosses the x-axis at a zero of odd multiplicity. In the section, we deliberately excluded the function f(x)=x from the discussion of the end behavior of f(x)=xn for odd numbers n and we said at the time that it was due to the fact that f(x)=x didn’t fit the pattern we were trying to establish. You just showed in the previous exercise that the end behavior of a linear function behaves like every other polynomial of odd degree, so what doesn’t f(x)=x do that g(x)=x3 does? It’s the ‘flattening’ for values of x near zero. It is this local behavior that will distinguish between a zero of multiplicity 1 and one of higher odd multiplicity. Look again closely at the graphs of a(x)=x(x+2)2 and F(x)=x3(x+2)2 from Exercise 3.1.1. Discuss with your classmates how the graphs are fundamentally different at the origin. It might help to use graphing technology to zoom in on the origin to see the different crossing behavior. Also compare the behavior of a(x)=x(x+2)2 to that of g(x)=x(x+2)3 near the point (2,0). What do you predict will happen at the zeros of f(x)=(x1)(x2)2(x3)3(x4)4(x5)5?
  10. Here are a few other questions for you to discuss with your classmates.
    1. How many local extrema could a polynomial of degree n have? How few local extrema can it have?
    2. Could a polynomial have two local maxima but no local minima?
    3. If a polynomial has two local maxima and two local minima, can it be of odd degree? Can it be of even degree?
    4. Can a polynomial have local extrema without having any real zeros?
    5. Why must every polynomial of odd degree have at least one real zero?
    6. Can a polynomial have two distinct real zeros and no local extrema?
    7. Can an x-intercept yield a local extrema? Can it yield an absolute extrema?
    8. If the y-intercept yields an absolute minimum, what can we say about the degree of the polynomial and the sign of the leading coefficient?

Answers

  1. f(x)=4x3x2
    Degree 2
    Leading term 3x2
    Leading coefficient 3
    Constant term 4
    As x,f(x)
    As x,f(x)

  2. g(x)=3x52x2+x+1
    Degree 5
    Leading term 3x5
    Leading coefficient 3
    Constant term 1
    As x,g(x)
    As x,g(x)

  3. q(r)=116r4
    Degree 4
    Leading term 16r4
    Leading coefficient 16
    Constant term 1
    As r,q(r)
    As r,q(r)

  4. Z(b)=42bb3
    Degree 3
    Leading term b3
    Leading coefficient 1
    Constant term 0
    As b,Z(b)
    As b,Z(b)

  5. f(x)=3x17+22.5x10πx7+13
    Degree 17
    Leading term 3x17
    Leading coefficient 3
    Constant term 13
    As x,f(x)
    As x,f(x)

  6. s(t)=4.9t2+v0t+s0
    Degree 2
    Leading term 4.9t2
    Leading coefficient 4.9
    Constant term s0
    As t,s(t)
    As t,s(t)

  7. P(x)=(x1)(x2)(x3)(x4)
    Degree 4
    Leading term x4
    Leading coefficient 1
    Constant term 24
    As x,P(x)
    As x,P(x)

  8. p(t)=t2(35t)(t2+t+4)
    Degree 5
    Leading term 5t5
    Leading coefficient 5
    Constant term 0
    As t,p(t)
    As t,p(t)

  9. f(x)=2x3(x+1)(x+2)2
    Degree 6
    Leading term 2x6
    Leading coefficient 2
    Constant term 0
    As x,f(x)
    As x,f(x)

  10. G(t)=4(t2)2(t+12)
    Degree 3
    Leading term 4t3
    Leading coefficient 4
    Constant term 8
    As t,G(t)
    As t,G(t)

  11. a(x)=x(x+2)2
    x=0 multiplicity 1
    x=2 multiplicity 2

    Screen Shot 2022-03-28 at 4.25.46 PM.png

  12. g(x)=x(x+2)3
    x=0 multiplicity 1
    x=2 multiplicity 3

    Screen Shot 2022-03-28 at 4.26.16 PM.png

  13. f(x)=2(x2)2(x+1)
    x=2 multiplicity 2
    x=1 multiplicity 1

    Screen Shot 2022-03-28 at 4.26.51 PM.png

  14. g(x)=(2x+1)2(x3)
    x=12 multiplicity 2
    x=3 multiplicity 1

    Screen Shot 2022-03-28 at 4.28.04 PM.png

  15. F(x)=x3(x+2)2
    x=0 multiplicity 3
    x=2 multiplicity 2

    Screen Shot 2022-03-28 at 4.28.38 PM.png

  16. P(x)=(x1)(x2)(x3)(x4)
    x=1 multiplicity 1
    x=2 multiplicity 1
    x=3 multiplicity 1
    x=4 multiplicity 1

    Screen Shot 2022-03-28 at 4.30.59 PM.png

  17. Q(x)=(x+5)2(x3)4
    x=5 multiplicity 2
    x=3 multiplicity 4

    Screen Shot 2022-03-28 at 4.31.28 PM.png

  18. f(x)=x2(x2)2(x+2)2
    x=2 multiplicity 2
    x=0 multiplicity 2
    x=2 multiplicity 2

    Screen Shot 2022-03-28 at 4.33.03 PM.png

  19. H(t)=(3t)(t2+1)
    x=3 multiplicity 1

    Screen Shot 2022-03-28 at 4.33.36 PM.png

  20. Z(b)=b(42b2)
    b=42 multiplicity 1
    b=0 multiplicity 1
    b=42 multiplicity 1

    Screen Shot 2022-03-28 at 4.34.07 PM.png

  21. g(x)=(x+2)3+1
    domain: (,)
    range: (,)

    Screen Shot 2022-03-28 at 4.34.45 PM.png

  22. g(x)=(x+2)4+1
    domain: (,)
    range: [1,)

    Screen Shot 2022-03-28 at 4.35.18 PM.png

  23. g(x)=23(x1)4
    domain: (,)
    range: (,2]

    Screen Shot 2022-03-28 at 4.48.40 PM.png

  24. g(x)=x53
    domain: (,)
    range: (,)

    Screen Shot 2022-03-28 at 4.49.23 PM.png

  25. g(x)=(x+1)5+10
    domain: (,)
    range: (,)

    Screen Shot 2022-03-28 at 4.50.01 PM.png

  26. g(x)=8x6
    domain: (,)
    range: (,8]

    Screen Shot 2022-03-28 at 4.54.08 PM.png

  27. We have f(4)=23,f(3)=5,f(0)=5,f(1)=3,f(2)=5 and f(3)=5 so the Intermediate Value Theorem tells us that f(x)=x39x+5 has real zeros in the intervals [4,3],[0,1] and [2,3].
  28. V(x)=x(8.52x)(112x)=4x339x2+93.5x, 0<x<4.25. Volume is maximized when x1.58, so the dimensions of the box with maximum volume are: height 1.58 inches, width 5.34 inches, and depth 7.84 inches. The maximum volume is 66.15 cubic inches.
  29. The calculator gives the location of the absolute maximum (rounded to three decimal places) as x6.305 and y1115.417. Since x represents the number of TVs sold in hundreds, x=6.305 corresponds to 630.5 TVs. Since we can’t sell half of a TV, we compare R(6.30)1115.415 and R(6.31)1115.416, so selling 631 TVs results in a (slightly) higher revenue. Since y represents the revenue in thousands of dollars, the maximum revenue is $1,115,416.
  30. P(x)=R(x)C(x)=5x3+35x245x25, 0x10.07.
  31. The calculator gives the location of the absolute maximum (rounded to three decimal places) as x3.897 and y35.255. Since x represents the number of TVs sold in hundreds, x=3.897 corresponds to 389.7 TVs. Since we can’t sell 0.7 of a TV, we compare P(3.89)35.254 and P(3.90)35.255, so selling 390 TVs results in a (slightly) higher revenue. Since y represents the revenue in thousands of dollars, the maximum revenue is $35,255.
  32. Making and selling 71 PortaBoys yields a maximized profit of $5910.67.
    1. Our ultimate goal is to maximize the volume, so we’ll start with the maximum Length + Girth of 130. This means the length is 1304x. The volume of a rectangular box is always length × width × height so we get V(x)=x2(1304x)=4x3+130x2.
    2. Graphing y=V(x) on [0,33]×[0,21000] shows a maximum at (21.67,20342.59) so the dimensions of the box with maximum volume are 21.67in.×21.67in.×43.32in. for a volume of 20342.59in.3.
    3. If we start with Length + Girth =108 then the length is 1084x and the volume is V(x)=4x3+108x2. Graphing y=V(x) on [0,27]×[0,11700] shows a maximum at (18.00,11664.00) so the dimensions of the box with maximum volume are 18.00in.×18.00in.×36in. for a volume of 11664.00in.3. (Calculus will confirm that the measurements which maximize the volume are 18in. by 18in. by 36in., however, as I’m sure you are aware by now, we treat all calculator results as approximations and list them as such.)
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  34. [OMITTED]

3.1E: Exercises is shared under a not declared license and was authored, remixed, and/or curated by LibreTexts.

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