Preliminaries: Real Numbers, Exponents, Factoring, and Analytic Geometry

Real Number System and Properties

  • Real numbers comprise all numbers positioned on the real number line, encompassing both rational and irrational numbers.

  • Classifications of Real Numbers:

    • Natural Numbers: The set of positive counting numbers starting from 11:     {1,2,3,4,… }\{1, 2, 3, 4, \dots\}
    • Integers: The set containing whole numbers, zero, and negative whole numbers:     {…,−3,−2,−1,0,1,2,3,… }\{\dots, -3, -2, -1, 0, 1, 2, 3, \dots\}
    • Rational Numbers: Numbers that can be written as a ratio of two integers in the form ab\frac{a}{b} where b≠0b \neq 0.
    • Examples: 12\frac{1}{2}, −73-\frac{7}{3}
    • Irrational Numbers: Numbers that cannot be expressed as a ratio of two integers. Their decimal representations are non-terminating and non-repeating.
    • Examples: 2\sqrt{2}, π\pi
  • Properties of Real Numbers:

    • Commutative Property of Addition: The order of addition does not affect the sum:     a+b=b+aa + b = b + a
    • Commutative Property of Multiplication: The order of multiplication does not affect the product:     a⋅b=b⋅aa \cdot b = b \cdot a
    • Associative Property of Addition: The grouping of terms does not affect the sum:     (a+b)+c=a+(b+c)(a + b) + c = a + (b + c)
    • Associative Property of Multiplication: The grouping of factors does not affect the product:     (a⋅b)⋅c=a⋅(b⋅c)(a \cdot b) \cdot c = a \cdot (b \cdot c)
    • Distributive Property: Multiplication distributes over addition:     a⋅(b+c)=a⋅b+a⋅ca \cdot (b + c) = a \cdot b + a \cdot c

Finite and Infinite Intervals

  • An interval represents a continuous subset of real numbers located on the number line.

  • Finite Intervals:

    • Open Interval (a,b)(a, b): Defined as {x∣a<x<b}\{x \mid a < x < b\}. Endpoints aa and bb are excluded.
    • Example: (−2,1)(-2, 1)
    • Closed Interval [a,b][a, b]: Defined as {x∣a≤x≤b}\{x \mid a \le x \le b\}. Endpoints aa and bb are included.
    • Example: [−1,2][-1, 2]
    • Half-Open Interval (a,b](a, b]: Defined as {x∣a<x≤b}\{x \mid a < x \le b\}. Endpoint aa is excluded; endpoint bb is included.
    • Example: (12,3]\left(\frac{1}{2}, 3\right]
    • Half-Open Interval [a,b)[a, b): Defined as {x∣a≤x<b}\{x \mid a \le x < b\}. Endpoint aa is included; endpoint bb is excluded.
    • Example: [−12,3)\left[-\frac{1}{2}, 3\right)

Finite Intervals Table

  • Infinite Intervals:
    • (a,∞)(a, \infty): Defined as {x∣x>a}\{x \mid x > a\}.
    • Example: (2,∞)(2, \infty)
    • [a,∞)[a, \infty): Defined as {x∣x≥a}\{x \mid x \ge a\}.
    • Example: [−1,∞)[-1, \infty)
    • (−∞,a)(-\infty, a): Defined as {x∣x<a}\{x \mid x < a\}.
    • Example: (−∞,1)(-\infty, 1)
    • (−∞,a](-\infty, a]: Defined as {x∣x≤a}\{x \mid x \le a\}.
    • Example: (−∞,−12]\left(-\infty, -\frac{1}{2}\right]

Infinite Intervals Table

Rules and Laws of Exponents

  • Definitions for ana^n (where base a>0a > 0):

    • Integer Exponent: If nn is a positive integer, then ana^n represents nn factors of aa multiplied together:     an=a⋅a⋅a⋅⋯⋅aa^n = a \cdot a \cdot a \cdot \dots \cdot a
    • Example: 25=2⋅2⋅2⋅2⋅2=322^5 = 2 \cdot 2 \cdot 2 \cdot 2 \cdot 2 = 32
    • Zero Exponent: Any non-zero base raised to the power of zero equals 11:     a0=1a^0 = 1
    • Example: 70=17^0 = 1
    • Exception: 000^0 is not defined.
    • Negative Exponent: If nn is a positive integer and a≠0a \neq 0:     a−n=1ana^{-n} = \frac{1}{a^n}
    • Example: 6−2=162=1366^{-2} = \frac{1}{6^2} = \frac{1}{36}
    • Fractional Exponents:
    • a. For a positive integer nn, a1/na^{1/n} or an\sqrt[n]{a} denotes the nthn\text{th} root of aa.
      • Example: 161/2=16=416^{1/2} = \sqrt{16} = 4
    • b. For positive integers mm and nn:       am/n=amn=(an)ma^{m/n} = \sqrt[n]{a^m} = (\sqrt[n]{a})^m
      • Example: 82/3=(83)2=(2)2=48^{2/3} = (\sqrt[3]{8})^2 = (2)^2 = 4
    • c. For positive integers mm and nn (with a≠0a \neq 0):       a−m/n=1am/na^{-m/n} = \frac{1}{a^{m/n}}
      • Example: 9−3/2=193/2=1(9)3=133=1279^{-3/2} = \frac{1}{9^{3/2}} = \frac{1}{(\sqrt{9})^3} = \frac{1}{3^3} = \frac{1}{27}
  • Laws of Exponents:

    1. Product Rule: am⋅an=am+na^m \cdot a^n = a^{m+n}
    • Example: x2⋅x3=x2+3=x5x^2 \cdot x^3 = x^{2+3} = x^5
    1. Quotient Rule: aman=am−n\frac{a^m}{a^n} = a^{m-n} (where a≠0a \neq 0)
    • Example: x7x4=x7−4=x3\frac{x^7}{x^4} = x^{7-4} = x^3
    1. Power of a Power Rule: (am)n=am⋅n(a^m)^n = a^{m \cdot n}
    • Example: (x4)3=x4⋅3=x12(x^4)^3 = x^{4 \cdot 3} = x^{12}
    1. Power of a Product Rule: (ab)n=an⋅bn(ab)^n = a^n \cdot b^n
    • Example: (2x)4=24⋅x4=16x4(2x)^4 = 2^4 \cdot x^4 = 16x^4
    1. Power of a Quotient Rule: (ab)n=anbn\left(\frac{a}{b}\right)^n = \frac{a^n}{b^n} (where b≠0b \neq 0)
    • Example: (x2)3=x323=x38\left(\frac{x}{2}\right)^3 = \frac{x^3}{2^3} = \frac{x^3}{8}
  • Common Pitfalls:

    • Avoid misapplying the product rule to powers of powers: (x2)3≠x5(x^2)^3 \neq x^5.
    • The correct expression multiplies the exponents: (x2)3=x2⋅3=x6(x^2)^3 = x^{2 \cdot 3} = x^6

Algebraic Expressions and Polynomials

  • Algebraic Expressions in Calculus:

    • Expressions commonly encountered include:     2x4/3−x1/3+12x^{4/3} - x^{1/3} + 12x2−x−2x2x^2 - x - \frac{2}{\sqrt{x}}3xy+2x+1\frac{3xy + 2}{x + 1}2x3+2x+12x^3 + 2x + 1
  • Monomials and Polynomials:

    • Monomial: An algebraic expression of the form axnax^n, where the coefficient aa is a real number and nn is a nonnegative integer. A monomial contains exactly one term.
    • Example: 7x27x^2
    • Polynomial: A monomial or the sum of two or more monomials.
    • Examples of polynomials:       x2+4x+4x^2 + 4x + 4x3+5x^3 + 5x4+3x2+3x^4 + 3x^2 + 3x2y+xy+yx^2 y + xy + y
  • Like (Similar) Terms:

    • Constant terms and terms that share the exact same variable factors are classified as like terms.
    • Like terms can be combined by adding or subtracting their numerical coefficients.
    • Examples:     3x+7x=10x3x + 7x = 10x12xy+3xy=72xy\frac{1}{2}xy + 3xy = \frac{7}{2}xy
    • This operation is mathematically justified by the distributive property of real numbers: ab+ac=a(b+c)ab + ac = a(b + c).

Factoring Algebraic Expressions

  • Factoring is the procedure of writing an algebraic expression as a product of simpler algebraic expressions.

  • Factoring Out the Greatest Common Term (GCF):

    • Always inspect an algebraic expression to check for common factors across all terms before applying other techniques.
    • Example 1: 3x2−x=x(3x−1)3x^2 - x = x(3x - 1)
    • Example 2: Factor 2a2x+4ax+6a2a^2 x + 4ax + 6a. The common factor across all terms is 2a2a:     2a2x+4ax+6a=2a⋅ax+2a⋅2x+2a⋅3=2a(ax+2x+3)2a^2 x + 4ax + 6a = 2a \cdot ax + 2a \cdot 2x + 2a \cdot 3 = 2a(ax + 2x + 3)
  • Factoring Second-Degree Polynomials (px2+qx+rpx^2 + qx + r):

    • Second-degree polynomials with integral coefficients factor into (ax+b)(cx+d)(ax + b)(cx + d), provided:     ac=pac = pad+bc=qad + bc = qbd=rbd = r
    • A trial-and-error method is systematically used to find factors.
  • Detailed Factoring Procedure for x2−2x−3x^2 - 2x - 3:

    • Step 1: Because the coefficient of x2x^2 is 11, the first-degree terms in each factor must be xx:     (x)(x)(x \quad)(x \quad)
    • Step 2: Identify integer pairs whose product equals the constant term −3-3. The possible pairs are (−1,3)(-1, 3) and (1,−3)(1, -3), yielding candidate factor sets:     (x−1)(x+3)(x - 1)(x + 3)(x+1)(x−3)(x + 1)(x - 3)
    • Step 3: Evaluate the inner and outer products to determine which set yields the linear term coefficient of −2-2:
    • Testing (x−1)(x+3)(x - 1)(x + 3): Sum of outer and inner product coefficients is (−1)(1)+(1)(3)=−1+3=2(-1)(1) + (1)(3) = -1 + 3 = 2
    • Testing (x+1)(x−3)(x + 1)(x - 3): Sum of outer and inner product coefficients is (1)(1)+(1)(−3)=1−3=−2(1)(1) + (1)(-3) = 1 - 3 = -2
    • Conclusion: The correct factorization is:     x2−2x−3=(x+1)(x−3)x^2 - 2x - 3 = (x + 1)(x - 3)

Quadratic Equations and the Quadratic Formula

  • A quadratic equation written in standard form is given by:   ax2+bx+c=0ax^2 + bx + c = 0   where a≠0a \neq 0.

  • The Quadratic Formula:

    • The explicit solutions for any quadratic equation in standard form are calculated using:     x=−b±b2−4ac2ax = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}

Cartesian Coordinate System and Lines

  • Cartesian Coordinate System:
    • Coordinate geometry (analytic geometry) represents geometric figures using ordered numerical pairs.
    • Formed by two perpendicular number lines intersecting at origin O(0,0)O(0, 0):
    • Horizontal axis: x-axisx\text{-axis}
    • Vertical axis: y-axisy\text{-axis}
    • Any point PP in two-dimensional space is designated as an ordered pair P(x,y)P(x, y).

Cartesian Coordinate System

  • Distance Formula:

    • The straight-line distance dd between two points P1(x1,y1)P_1(x_1, y_1) and P2(x2,y2)P_2(x_2, y_2) in the Cartesian plane is:     d=(x2−x1)2+(y2−y1)2d = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}
  • Slope of a Nonvertical Line:

    • The slope mm measures the steepness and direction of line LL passing through points (x1,y1)(x_1, y_1) and (x2,y2)(x_2, y_2), where x1≠x2x_1 \neq x_2:     m=ΔyΔx=y2−y1x2−x1m = \frac{\Delta y}{\Delta x} = \frac{y_2 - y_1}{x_2 - x_1}

Slope Diagram

  • Summary of Equations of Straight Lines:
    • Vertical Line: x=ax = a
    • Horizontal Line: y=by = b
    • Point-Slope Form: y−y1=m(x−x1)y - y_1 = m(x - x_1)
    • Slope-Intercept Form: y=mx+by = mx + b (where mm is the slope and (0,b)(0, b) is the y-intercepty\text{-intercept})
    • General Form of a Linear Equation: Ax+By+C=0Ax + By + C = 0 (where AA, BB, and CC are constants, and AA and BB are not both zero)

Worked Examples

  • Example 0.1: Simplify the following exponent and radical expressions:

    • a. (3x2)(4x3)(3x^2)(4x^3)
    • Multiply coefficients and add exponents of matching bases:       (3⋅4)(x2+3)=12x5(3 \cdot 4)(x^{2+3}) = 12x^5
    • b. 165/4161/2\frac{16^{5/4}}{16^{1/2}}
    • Apply quotient law for exponents:       165/4−1/2=165/4−2/4=163/416^{5/4 - 1/2} = 16^{5/4 - 2/4} = 16^{3/4}
    • Simplify 163/416^{3/4}:       163/4=(164)3=(2)3=816^{3/4} = (\sqrt[4]{16})^3 = (2)^3 = 8
    • c. (62/3)3(6^{2/3})^3
    • Apply power of a power law:       6(2/3)⋅3=62=366^{(2/3) \cdot 3} = 6^2 = 36
    • d. (x3y−2)−2(x^3 y^{-2})^{-2}
    • Distribute exponent across factors:       x3(−2)y−2(−2)=x−6y4=y4x6x^{3(-2)} y^{-2(-2)} = x^{-6} y^4 = \frac{y^4}{x^6}
    • e. (y3/2x1/4)−2\left(\frac{y^{3/2}}{x^{1/4}}\right)^{-2}
    • Apply negative power to numerator and denominator:       y(3/2)(−2)x(1/4)(−2)=y−3x−1/2=x1/2y3=xy3\frac{y^{(3/2)(-2)}}{x^{(1/4)(-2)}} = \frac{y^{-3}}{x^{-1/2}} = \frac{x^{1/2}}{y^3} = \frac{\sqrt{x}}{y^3}
  • Example 0.2: Factorize the following algebraic expressions:

    • a. 2ax+2ay+bx+by2ax + 2ay + bx + by
    • Group terms into pairs:       (2ax+2ay)+(bx+by)(2ax + 2ay) + (bx + by)
    • Factor out common terms from each group:       2a(x+y)+b(x+y)2a(x + y) + b(x + y)
    • Factor out binomial (x+y)(x + y):       (2a+b)(x+y)(2a + b)(x + y)
    • b. 3xy−4−2y+6x3x\sqrt{y} - 4 - 2\sqrt{y} + 6x
    • Rearrange terms to group common factors:       (3xy+6x)−(2y+4)(3x\sqrt{y} + 6x) - (2\sqrt{y} + 4)
    • Factor out common terms:       3x(y+2)−2(y+2)3x(\sqrt{y} + 2) - 2(\sqrt{y} + 2)
    • Factor out common binomial (y+2)(\sqrt{y} + 2):       (3x−2)(y+2)(3x - 2)(\sqrt{y} + 2)
  • Example 0.3: Factorize the following quadratic polynomials:

    • a. 3x2+4x−43x^2 + 4x - 4
    • Test factors of form (3x+b)(x+d)(3x + b)(x + d) where b⋅d=−4b \cdot d = -4:       (3x−2)(x+2)(3x - 2)(x + 2)
    • Check middle term: 3x(2)+(−2)(x)=6x−2x=4x3x(2) + (-2)(x) = 6x - 2x = 4x.
    • Correct factorization: (3x−2)(x+2)(3x - 2)(x + 2)
    • b. 3x2−6x−243x^2 - 6x - 24
    • Factor out greatest numerical common factor 33:       3(x2−2x−8)3(x^2 - 2x - 8)
    • Factor the remaining quadratic expression x2−2x−8x^2 - 2x - 8 into two binomials:       3(x−4)(x+2)3(x - 4)(x + 2)
  • Example 0.4: Solve each quadratic equation:

    • a. 2x2+5x−12=02x^2 + 5x - 12 = 0
    • Identify coefficients: a=2a = 2, b=5b = 5, c=−12c = -12
    • Substitute into quadratic formula:       x=−5±52−4(2)(−12)2(2)x = \frac{-5 \pm \sqrt{5^2 - 4(2)(-12)}}{2(2)}x=−5±25+964=−5±1214=−5±114x = \frac{-5 \pm \sqrt{25 + 96}}{4} = \frac{-5 \pm \sqrt{121}}{4} = \frac{-5 \pm 11}{4}
    • Calculate solutions:       x1=−5+114=64=32x_1 = \frac{-5 + 11}{4} = \frac{6}{4} = \frac{3}{2}x2=−5−114=−164=−4x_2 = \frac{-5 - 11}{4} = \frac{-16}{4} = -4
    • b. x2=−3x+8x^2 = -3x + 8
    • Rearrange into standard form ax2+bx+c=0ax^2 + bx + c = 0:       x2+3x−8=0x^2 + 3x - 8 = 0
    • Identify coefficients: a=1a = 1, b=3b = 3, c=−8c = -8
    • Substitute into quadratic formula:       x=−3±32−4(1)(−8)2(1)x = \frac{-3 \pm \sqrt{3^2 - 4(1)(-8)}}{2(1)}x=−3±9+322=−3±412x = \frac{-3 \pm \sqrt{9 + 32}}{2} = \frac{-3 \pm \sqrt{41}}{2}
  • Example 0.5 (Applied Distance and Cost Problem):

Submarine Cable Route

  • Problem: Power relay station SS is positioned on a straight coastal highway at coordinates S(10000,0)S(10000, 0) (in feet). A marine biology experimental station MM is located on an island at coordinates M(0,3000)M(0, 3000) (in feet). A cable is laid from S(10000,0)S(10000, 0) along land to point Q(2000,0)Q(2000, 0), and then under water from Q(2000,0)Q(2000, 0) directly to M(0,3000)M(0, 3000). Land cable costs $3.00 per running foot\$3.00\text{ per running foot}, while underwater cable costs $5.00 per running foot\$5.00\text{ per running foot}. Calculate the total cost for laying the cable.

  • Step 1: Determine the length of land cable from S(10000,0)S(10000, 0) to Q(2000,0)Q(2000, 0):     dland=10000−2000=8000 feetd_{\text{land}} = 10000 - 2000 = 8000\,\text{feet}

  • Step 2: Determine the length of underwater cable from Q(2000,0)Q(2000, 0) to M(0,3000)M(0, 3000) using the distance formula:     dwater=(0−2000)2+(3000−0)2=(−2000)2+(3000)2d_{\text{water}} = \sqrt{(0 - 2000)^2 + (3000 - 0)^2} = \sqrt{(-2000)^2 + (3000)^2}dwater=4,000,000+9,000,000=13,000,000=100013 feet≈3605.551 feetd_{\text{water}} = \sqrt{4,000,000 + 9,000,000} = \sqrt{13,000,000} = 1000\sqrt{13}\,\text{feet} \approx 3605.551\,\text{feet}

  • Step 3: Calculate total cable installation cost:     Costland=8000⋅3.00=$24,000.00\text{Cost}_{\text{land}} = 8000 \cdot 3.00 = \$24,000.00Costwater=100013⋅5.00=500013≈$18,027.76\text{Cost}_{\text{water}} = 1000\sqrt{13} \cdot 5.00 = 5000\sqrt{13} \approx \$18,027.76Total Cost=24,000+18,027.76=$42,027.76\text{Total Cost} = 24,000 + 18,027.76 = \$42,027.76

    • Example 0.6: Find an equation of the line passing through point (1,3)(1, 3) with slope m=2m = 2
  • Use point-slope form y−y1=m(x−x1)y - y_1 = m(x - x_1):     y−3=2(x−1)y - 3 = 2(x - 1)y−3=2x−2y - 3 = 2x - 2y=2x+1y = 2x + 1

    • Example 0.7: Find an equation of the line passing through points (−3,2)(-3, 2) and (4,−1)(4, -1)
  • Step 1: Calculate slope mm:     m=−1−24−(−3)=−37=−37m = \frac{-1 - 2}{4 - (-3)} = \frac{-3}{7} = -\frac{3}{7}

  • Step 2: Apply point-slope form using point (−3,2)(-3, 2):     y−2=−37(x−(−3))y - 2 = -\frac{3}{7}(x - (-3))y−2=−37(x+3)y - 2 = -\frac{3}{7}(x + 3)y−2=−37x−97y - 2 = -\frac{3}{7}x - \frac{9}{7}y=−37x+57y = -\frac{3}{7}x + \frac{5}{7}(In general form: 3x+7y−5=03x + 7y - 5 = 0)

    • Example 0.8 (Applied Linear Sales Prediction):

Sales Trend Graph

  • Problem: A sales manager plotted annual store sales versus time for the past 5 years and noted that the points lie approximately along a straight line. Using data points corresponding to Year 1 (20 thousand dollars20\text{ thousand dollars}) and Year 5 (60 thousand dollars60\text{ thousand dollars}), establish the equation of the trend line and predict sales for Year 6.
  • Step 1: Identify coordinate points: (1,20)(1, 20) and (5,60)(5, 60).
  • Step 2: Compute slope mm:     m=60−205−1=404=10 thousand dollars per yearm = \frac{60 - 20}{5 - 1} = \frac{40}{4} = 10\,\text{thousand dollars per year}
  • Step 3: Determine line equation using point (1,20)(1, 20):     y−20=10(x−1)y - 20 = 10(x - 1)y−20=10x−10y - 20 = 10x - 10y=10x+10y = 10x + 10
  • Step 4: Predict sales for Year 6 (x=6x = 6):     y=10(6)+10=60+10=70 thousand dollarsy = 10(6) + 10 = 60 + 10 = 70\,\text{thousand dollars}
  • The predicted sales figure for the sixth year is $70,000\$70,000.