Rational Expressions, Applications, and Integral Exponents — Study Notes

Rational Expressions: LCD and Adding/Subtracting

  • Core idea: To add or subtract rational expressions, you must have a common denominator. The best common denominator (LCD) is the Least Common Denominator, which is the product of all distinct factors appearing in the denominators, each raised to the highest power it occurs with.

  • Quick reminder: If two fractions already share a common denominator, you can add/subtract directly. The LCD is just a preferred common denominator that makes simplification easier later.

  • Procedure to find the LCD (general steps):

    • Factor each denominator completely.
    • For the LCD, take each distinct factor to the highest power it appears in any denominator.
    • Write each fraction with the LCD by multiplying its numerator and denominator by any missing factors.
    • Combine the fractions over the LCD and simplify.

  • Example 1 (two fractions):

    • Problem: compute xx−3+x−1x+3\frac{x}{x-3} + \frac{x-1}{x+3}
    • Denominators: $(x-3)$ and $(x+3)$; LCD is (x−3)(x+3)=x2−9.(x-3)(x+3) = x^2-9.
    • Rewrite each fraction with the LCD:
    • xx−3=x(x+3)(x−3)(x+3)=x2+3x(x−3)(x+3)\frac{x}{x-3} = \frac{x(x+3)}{(x-3)(x+3)} = \frac{x^2+3x}{(x-3)(x+3)}
    • x−1x+3=(x−3)(x−1)(x−3)(x+3)=x2−4x+3(x−3)(x+3)\frac{x-1}{x+3} = \frac{(x-3)(x-1)}{(x-3)(x+3)} = \frac{x^2-4x+3}{(x-3)(x+3)}
    • Add numerators over the common denominator:
    • Numerator: (x2+3x)+(x2−4x+3)=2x2−x+3(x^2+3x) + (x^2-4x+3) = 2x^2 - x + 3
    • Result: 2x2−x+3(x−3)(x+3)\frac{2x^2 - x + 3}{(x-3)(x+3)}
    • Note: The numerator does not factor nicely to cancel with the denominator in this case.

  • Example 2 (three fractions):

    • Problem: m−3m−2+7−4m2m2−9m+10+m+12m−5\frac{m-3}{m-2} +\frac{7-4m}{2m^2-9m+10} +\frac{m+1}{2m-5}
    • Factor the denominator of the middle fraction: 2m2−9m+10=(2m−5)(m−2)2m^2-9m+10 = (2m-5)(m-2)
    • LCD: (2m−5)(m−2)(2m-5)(m-2)
    • Rewrite each fraction with the LCD:
    • First: m−3m−2=(m−3)(2m−5)(2m−5)(m−2)\frac{m-3}{m-2} = \frac{(m-3)(2m-5)}{(2m-5)(m-2)}
    • Second: 7−4m(2m−5)(m−2)\frac{7-4m}{(2m-5)(m-2)} (already over the LCD)
    • Third: m+12m−5=(m+1)(m−2)(2m−5)(m−2)\frac{m+1}{2m-5} = \frac{(m+1)(m-2)}{(2m-5)(m-2)}
    • Compute numerators over the common denominator:
    • $(m-3)(2m-5) = 2m^2 - 11m + 15$
    • Numerator of middle term: $(7-4m)$
    • $(m+1)(m-2) = m^2 - m - 2$
    • Sum numerators:
    • (2m2−11m+15)+(7−4m)+(m2−m−2)=3m2−16m+20(2m^2 - 11m + 15) + (7 - 4m) + (m^2 - m - 2) = 3m^2 - 16m + 20
    • Numerator factors: 3m2−16m+20=(m−2)(3m−10)3m^2 - 16m + 20 = (m-2)(3m-10)
    • Cancel common factor $(m-2)$ with the denominator to obtain:
    • Final: 3m−102m−5\frac{3m-10}{2m-5}
    • Domain note: denominators cannot be zero, so $m \neq 2$ and $m \neq \frac{5}{2}$.

  • Key takeaways and tips

    • Always check if denominators share a common factor; cancel only after combining over the LCD.
    • If the resulting numerator factors, you may be able to cancel with the LCD to simplify further.
    • If a factor appears in more than one denominator, include it only to the highest power it appears with.
    • Domain restrictions: keep track of where denominators are zero to avoid extraneous or undefined results.

Applications: Revenue, Cost, Profit, and Average Cost

  • Important definitions:
    • Revenue (R): revenue is price per item times number of items sold:
      R(x)=(extpriceperitem)×x.R(x) = ( ext{price per item})\times x.
    • Total cost (C): split into fixed cost and variable cost:
      C(x)=extFixedcost+extVariablecost(peritem)×x.C(x) = ext{Fixed cost} + ext{Variable cost (per item)}\times x.
    • Fixed cost: a cost that does not depend on quantity produced; constant.
    • Variable cost: cost that varies with the number of items produced.
    • Profit (P):
      P(x)=R(x)−C(x).P(x) = R(x) - C(x).
    • Average cost (\bar{C}(x)):
      Cˉ(x)=C(x)x.\bar{C}(x) = \frac{C(x)}{x}.
  • Worked example (company manufacturing guitars):
    • Given:
    • Fixed monthly cost: $95,000\$95{,}000
    • Variable cost per guitar: $25\$25
    • Price per guitar: $75\$75
    • Maximum production per month: 10{,}000 units
    • All manufactured units are sold.
    • Total cost as a function of x guitars:
      C(x)=95,000+25x.C(x) = 95{,}000 + 25x.
    • Revenue as a function of x guitars:
      R(x)=75x.R(x) = 75x.
    • Profit function:
      P(x)=R(x)−C(x)=75x−(95,000+25x)=50x−95,000.P(x) = R(x) - C(x) = 75x - (95{,}000 + 25x) = 50x - 95{,}000.
    • Domain restriction (feasibility): 0≤x≤10,000.0 \le x \le 10{,}000.
    • Average cost function:
      Cˉ(x)=C(x)x=95,000+25xx=95,000x+25.\bar{C}(x) = \frac{C(x)}{x} = \frac{95{,}000 + 25x}{x} = \frac{95{,}000}{x} + 25.
    • Interpretations of examples:
    • If the company makes 1{,}000 guitars in a month, the average cost per guitar is:
      Cˉ(1000)=95,0001000+25=95+25=$120.\bar{C}(1000) = \frac{95{,}000}{1000} + 25 = 95 + 25 = \$120.
    • If the company makes 5{,}000 guitars in a month, the average cost per guitar is:
      Cˉ(5000)=95,0005000+25=19+25=$44.\bar{C}(5000) = \frac{95{,}000}{5000} + 25 = 19 + 25 = \$44.
    • Practical interpretation: average cost is the per-unit average across all units produced; it decreases as production increases (due to spreading fixed costs over more units).

Section 1.5: Integral Exponents

  • Recap of exponent basics:
    • Positive integer exponents: a^n means multiply a by itself n times.
    • Exponent zero rule (for nonzero a): a^0 = 1.
    • 0^0 is undefined (as discussed in class).
    • Convention on how exponent zero applies when expressions have multiple factors or signs is important (see examples in lecture for clarification).
  • Negative exponent rule:
    • For nonzero a and positive integer n:
      a−n=1an.a^{-n} = \frac{1}{a^{n}}.
    • Example: 7−2=172=149.7^{-2} = \frac{1}{7^{2}} = \frac{1}{49}.
    • Example: (−2)−5=1(−2)5=1−32=−132.(-2)^{-5} = \frac{1}{(-2)^{5}} = \frac{1}{-32} = -\frac{1}{32}.
    • If a negative exponent appears in the denominator, you can move it to the numerator to obtain a positive exponent, and move the sign accordingly:
      6−2=162=136.6^{-2} = \frac{1}{6^{2}} = \frac{1}{36}.
  • Power to a power (the nested exponent rule):
    • (am)n=amn.(a^m)^n = a^{mn}.
    • Examples:
    • (x−2)−3=x(−2)⋅(−3)=x6.(x^{-2})^{-3} = x^{(-2)\cdot(-3)} = x^{6}.
    • (y5)4=y20.(y^5)^4 = y^{20}.
  • Product to a power:
    • If you raise a product to a power, raise each factor to that power:
      (ab)n=anbn.(ab)^n = a^n b^n.
    • Examples:
    • ( (3x)^4 = 3^4 x^4 = 81 x^4 ).
    • ( (-5 y^4)^2 = (-5)^2 y^{8} = 25 y^8 ).
    • For a product with multiple factors, apply the rule to each factor.
  • A more complex example combining rules:
    • Consider ( (-2 x^{-2} y^4)^3 ):
    • Apply product-to-powers:
      [ (-2)^3 (x^{-2})^3 (y^4)^3 = -8 \cdot x^{-6} \cdot y^{12}. ]\
    • To remove negative exponents:
      [ -8 \cdot \frac{y^{12}}{x^{6}} = -\frac{8 y^{12}}{x^{6}}. ]\
    • If you prefer, move the negative exponents to the numerator: same form (the sign can be placed in front of the fraction).
  • Quotient rule (power rule for division):
    • If bases are equal, subtract exponents:
      aman=am−n,a≠0.\frac{a^m}{a^n} = a^{m-n},\quad a \neq 0.
    • Examples:
    • x11x5=x11−5=x6.\frac{x^{11}}{x^5} = x^{11-5} = x^{6}.
    • (5x)8/(5x)3=(5x)8−3=(5x)5=55x5=3125x5.(5x)^8 / (5x)^3 = (5x)^{8-3} = (5x)^5 = 5^5 x^5 = 3125 x^5.
    • z10z−6=z10−(−6)=z16.\frac{z^{10}}{z^{-6}} = z^{10 - (-6)} = z^{16}.
  • Quick practical tips:
    • Always check the base when using exponent rules; the rules apply to identical bases.
    • When simplifying expressions with negative exponents, convert to positive exponents by moving terms to the numerator/denominator as appropriate.
    • Keep track of domain restrictions (e.g., you cannot have a base of 0 with negative exponents).

Quick practice prompts and connections

  • Practice picking LCDs for given sets of denominators, rewriting each fraction, combining numerators, and simplifying.
  • Practice applying exponent rules to nested expressions, including products like (ab)^n, and quotients with common bases.
  • Real-world tie-ins: rate-based problems (revenue/cost) often lead to linear expressions in x; integrating exponent rules helps simplify algebraic expressions that appear in modeling.