In Exercises diffquotexerfirst - diffquotexerlast, find and simplify the difference quotient \(\dfrac {f(x+h) - f(x)}{h}\) for the given function.

\(f(x) = 2x - 5\)

\(2\)
\(f(x) = -3x + 5\)

\(-3\)
\(f(x) = 6\)

\(0\)
\(f(x) = 3x^2 - x\)

\(6x+3h-1\)
\(f(x) = -x^2 + 2x - 1\)

\(-2x-h+2\)
\(f(x) = 4x^2\)

\(8x+4h\)
\(f(x) = x-x^2\)

\(-2x-h+1\)
\(f(x) = x^{3} + 1\)

\(3x^{2} + 3xh + h^{2}\)
\(f(x) = mx + b\;\) where \(m \neq 0\)

\(m\)
\(f(x) = ax^{2} + bx + c\;\) where \(a \neq 0\)

\(2ax + ah + b\)
\(f(x) = \dfrac {2}{x}\)

\(\dfrac {-2}{x(x+h)}\)
\(f(x) = \dfrac {3}{1-x}\)

\(\dfrac {3}{(1-x-h)(1-x)}\)
\(f(x) = \dfrac {1}{x^2}\)

\(\dfrac {-(2x+h)}{x^2(x+h)^2}\)
\(f(x) = \dfrac {2}{x+5}\)

\(\dfrac {-2}{(x+5)(x+h+5)}\)
\(f(x) = \dfrac {1}{4x-3}\)

\(\dfrac {-4}{(4x-3)(4x+4h-3)}\)
\(f(x) = \dfrac {3x}{x+1}\)

\(\dfrac {3}{(x+1)(x+h+1)}\)
\(f(x) = \dfrac {x}{x - 9}\)

\(\dfrac {-9}{(x - 9)(x + h - 9)}\)
\(f(x) = \dfrac {x^2}{2x+1}\)

\(\dfrac {2x^2+2xh+2x+h}{(2x+1)(2x+2h+1)}\)
\(f(x) = \sqrt {x-9}\)

\(\dfrac {1}{\sqrt {x+h-9} + \sqrt {x-9}}\)
\(f(x) = \sqrt {2x+1}\)

\(\dfrac {2}{\sqrt {2x+2h+1} + \sqrt {2x+1}}\)
\(f(x) = \sqrt {-4x+5}\)

\(\dfrac {-4}{\sqrt {-4x-4h+5} + \sqrt {-4x+5}}\)
\(f(x) = \sqrt {4-x}\)

\(\dfrac {-1}{\sqrt {4-x-h} + \sqrt {4-x}}\)
\(f(x) = \sqrt {ax+b}\), where \(a \neq 0\).

\(\dfrac {a}{\sqrt {ax+ah+b} + \sqrt {ax+b}}\)
\(f(x) = x \sqrt {x}\)

\(\dfrac {3x^2+3xh+h^2}{(x+h)^{3/2} + x^{3/2}}\)
\(f(x) = \sqrt [3]{x}\)

\((a-b)\left (a^2+ab+b^2\right ) = a^3 - b^3\)

\(\dfrac {1}{(x+h)^{2/3} + (x+h)^{1/3} x^{1/3} + x^{2/3}}\)

In Exercises econexerfirst - econexerlast, \(C(x)\) denotes the cost to produce \(x\) items and \(p(x)\) denotes the price-demand function in the given economic scenario. In each Exercise, do the following:

  • Find and interpret \(C(0)\).
  • Find and interpret \(\overline {C}(10)\).
  • Find and interpret \(p(5)\)
  • Find and simplify \(R(x)\).
  • Find and simplify \(P(x)\).
  • Solve \(P(x) = 0\) and interpret.
The cost, in dollars, to produce \(x\) “I’d rather be a Sasquatch” T-Shirts is \(C(x) = 2x+26\), \(x \geq 0\) and the price-demand function, in dollars per shirt, is \(p(x) = 30 - 2x\), \(0 \leq x \leq 15\).

  • Find and interpret \(C(0)\).

    \(C(0) = 26\), so the fixed costs are \(\$26\).

  • Find and interpret \(\overline {C}(10)\).

    \(\overline {C}(10) = 4.6\), so when 10 shirts are produced, the cost per shirt is \(\$4.60\).

  • Find and interpret \(p(5)\)

    \(p(5) = 20\), so to sell \(5\) shirts, set the price at \(\$20\) per shirt.

  • Find and simplify \(R(x)\).

    \(R(x) = -2x^2+30x\), \(0 \leq x \leq 15\)

  • Find and simplify \(P(x)\).

    \(P(x) = -2x^2+28x-26\), \(0 \leq x \leq 15\)

  • Solve \(P(x) = 0\) and interpret.

    \(P(x) = 0\) when \(x = 1\) and \(x=13\). These are the ‘break even’ points, so selling \(1\) shirt or \(13\) shirts will guarantee the revenue earned exactly recoups the cost of production.
The cost, in dollars, to produce \(x\) bottles of \(100 \%\) All-Natural Certified Free-Trade Organic Sasquatch Tonic is \(C(x) = 10x+100\), \(x \geq 0\) and the price-demand function, in dollars per bottle, is \(p(x) = 35 - x\), \(0 \leq x \leq 35\).

  • Find and interpret \(C(0)\).

    \(C(0) = 100\), so the fixed costs are \(\$100\).

  • Find and interpret \(\overline {C}(10)\).

    \(\overline {C}(10) = 20\), so when 10 bottles of tonic are produced, the cost per bottle is \(\$20\).

  • Find and interpret \(p(5)\)

    \(p(5) = 30\), so to sell \(5\) bottles of tonic, set the price at \(\$30\) per bottle.

  • Find and simplify \(R(x)\).

    \(R(x) = -x^2+35x\), \(0 \leq x \leq 35\)

  • Find and simplify \(P(x)\).

    \(P(x) = -x^2+25x-100\), \(0 \leq x \leq 35\)

  • Solve \(P(x) = 0\) and interpret.

    \(P(x) = 0\) when \(x = 5\) and \(x=20\). These are the ‘break even’ points, so selling \(5\) bottles of tonic or \(20\) bottles of tonic will guarantee the revenue earned exactly recoups the cost of production.
The cost, in cents, to produce \(x\) cups of Mountain Thunder Lemonade at Junior’s Lemonade Stand is \(C(x) = 18x + 240\), \(x \geq 0\) and the price-demand function, in cents per cup, is \(p(x) = 90-3x\), \(0 \leq x \leq 30\).

  • Find and interpret \(C(0)\).

    \(C(0) = 240\), so the fixed costs are \(240\)¢ or \(\$2.40\).

  • Find and interpret \(\overline {C}(10)\).

    \(\overline {C}(10) = 42\), so when 10 cups of lemonade are made, the cost per cup is \(42\)¢.

  • Find and interpret \(p(5)\)

    \(p(5) = 75\), so to sell \(5\) cups of lemonade, set the price at \(75\)¢ per cup.

  • Find and simplify \(R(x)\).

    \(R(x) = -3x^2+90x\), \(0 \leq x \leq 30\)

  • Find and simplify \(P(x)\).

    \(P(x) = -3x^2+72x-240\), \(0 \leq x \leq 30\)

  • Solve \(P(x) = 0\) and interpret.

    \(P(x) = 0\) when \(x = 4\) and \(x=20\). These are the ‘break even’ points, so selling \(4\) cups of lemonade or \(20\) cups of lemonade will guarantee the revenue earned exactly recoups the cost of production.
The daily cost, in dollars, to produce \(x\) Sasquatch Berry Pies \(C(x) = 3x + 36\), \(x \geq 0\) and the price-demand function, in dollars per pie, is \(p(x) = 12-0.5x\), \(0 \leq x \leq 24\).

  • Find and interpret \(C(0)\).

    \(C(0) = 36\), so the daily fixed costs are \(\$36\).

  • Find and interpret \(\overline {C}(10)\).

    \(\overline {C}(10) = 6.6\), so when 10 pies are made, the cost per pie is \(\$6.60\).

  • Find and interpret \(p(5)\)

    \(p(5) = 9.5\), so to sell \(5\) pies a day, set the price at \(\$9.50\) per pie.

  • Find and simplify \(R(x)\).

    \(R(x) = -0.5 x^2 + 12x\), \(0 \leq x \leq 24\)

  • Find and simplify \(P(x)\).

    \(P(x) = -0.5 x^2+9x-36\), \(0 \leq x \leq 24\)

  • Solve \(P(x) = 0\) and interpret.

    \(P(x) = 0\) when \(x = 6\) and \(x=12\). These are the ‘break even’ points, so selling \(6\) pies or \(12\) pies a day will guarantee the revenue earned exactly recoups the cost of production.
The monthly cost, in hundreds of dollars, to produce \(x\) custom built electric scooters is \(C(x) = 20x + 1000\), \(x \geq 0\) and the price-demand function, in hundreds of dollars per scooter, is \(p(x) = 140-2x\), \(0 \leq x \leq 70\).

  • Find and interpret \(C(0)\).

    \(C(0) = 1000\), so the monthly fixed costs are \(1000\) hundred dollars, or \(\$100,\!000\).

  • Find and interpret \(\overline {C}(10)\).

    \(\overline {C}(10) = 120\), so when 10 scooters are made, the cost per scooter is \(120\) hundred dollars, or \(\$12,\!000\).

  • Find and interpret \(p(5)\)

    \(p(5) = 130\), so to sell \(5\) scooters a month, set the price at \(130\) hundred dollars, or \(\$13,\!000\) per scooter.

  • Find and simplify \(R(x)\).

    \(R(x) = -2x^2+140x\), \(0 \leq x \leq 70\)

  • Find and simplify \(P(x)\).

    \(P(x) = -2x^2+120x-1000\), \(0 \leq x \leq 70\)

  • Solve \(P(x) = 0\) and interpret.

    \(P(x) = 0\) when \(x = 10\) and \(x=50\). These are the ‘break even’ points, so selling \(10\) scooters or \(50\) scooters a month will guarantee the revenue earned exactly recoups the cost of production.

Let us return to Example marginalsetupex where \(C(x) = .03x^{3} - 4.5x^{2} + 225x + 250\) denotes the cost, in dollars, of producing \(x\) PortaBoy game systems. Recall the average cost is defined as \(\overline {C}(x) = \frac {C(x)}{x}\), \(x > 0\), is the cost per item.

  1. Find and interpret \(\overline {C}(75)\).
  2. Define the marginal cost \(MC(x) = C(x+1) - C(x)\). Find and interpret \(MC(75)\).
  3. How do your answers to parts ACexercise1 and MCexercise1 compare?
  4. Graph \(y = \overline {C}(x)\) with help from a graphing utility. What is happening graphically near \(x = 75\)?

  5. Use Theorem functionarithmeticaroc to show that, in general, \(\text {ARoC}[ \overline {C}(x)] = 0\) when \(MC(x) = \overline {C}(x)\).

    HINT: Note that, by definition, \(MC(x) = C(x+1) - C(x) = \Delta [C(x)]\) when \(\Delta x = 1\).

    Hence, \(\text {ARoC}[C(x)] = \frac {\Delta [C(x)]}{\Delta x} = \frac {\Delta [C(x)]}{1} = \Delta [C(x)] = MC(x)\) in this case …