In Exercises diffquotexerfirsta - diffquotexerlasta, find the limit of the following difference quotients.
  1. \(\lim _{h \rightarrow 0} \frac {f(2+h) - f(2)}{h}\)
  2. \(\lim _{h \rightarrow 0} \frac {f(x+h) - f(x)}{h}\)
\(f(x) = 2x - 5\)

  1. \(\lim _{h \rightarrow 0} 2 = 2\)
  2. \(\lim _{h \rightarrow 0} 2 = 2\)
\(f(x) = -3x + 5\)

  1. \(\lim _{h \rightarrow 0} -3 = -3\)
  2. \(\lim _{h \rightarrow 0} -3 = -3\)
\(f(x) = 6\)

  1. \(\lim _{h \rightarrow 0} 0 = 0\)
  2. \(\lim _{h \rightarrow 0} 0 = 0\)
\(f(x) = 3x^2 - x\)

  1. \(\lim _{h \rightarrow 0} (3h+11)= 11\)
  2. \(\lim _{h \rightarrow 0} (6x+3h-1)= 6x-1\)
\(f(x) = -x^2 + 2x - 1\)

  1. \(\lim _{h \rightarrow 0} (-h-2) = -2\)
  2. \(\lim _{h \rightarrow 0} (-2x-h+2) = -2x+2\)
\(f(x) = 4x^2\)

  1. \(\lim _{h \rightarrow 0} (4h+16) = 16\)
  2. \(\lim _{h \rightarrow 0} (8x+4h) = 8x\)
In Exercises tangentlinepolyfirst - tangentlinepolylast, find:
  1. \(f'(2) = \lim _{h \rightarrow 0} \frac {f(2+h) - f(2)}{h}\)
  2. The equation of the tangent line at \((2, f(2))\). Check your answer graphically.
  3. \(f'(x) = \lim _{h \rightarrow 0} \frac {f(x+h) - f(x)}{h}\)
  4. The equation of the tangent line at \((0,f(0))\). Check your answer graphically.
\(f(x) = x-x^2\)

  1. \(f'(2) = \lim _{h \rightarrow 0} (-h-3) = -3\)
  2. \(y = f'(2)(x-2) + f(2) = (-3)(x-2)+(-2)\) so \(y = -3x+4\).
  3. \(f'(x) = \lim _{h \rightarrow 0} (-2x-h+1) = -2x+1\)
  4. \(y = f'(0)(x-0) + f(0) = (1)(x-0) + 0\) so \(y = x\).
\(f(x) = x^{3} + 1\)

  1. \(f'(2) = \lim _{h \rightarrow 0} \left ( h^2+6h+12 \right ) = 12\)
  2. \(y = f'(2)(x-2) + f(2) = 12(x-2)+ 9\) so \(y = 12x - 15\).
  3. \(f'(x) = \lim _{h \rightarrow 0} \left (3x^{2} + 3xh + h^{2} \right ) = 3x^{2}\)
  4. \(y = f'(0)(x-0)+f(0) = 0(x-0)+1\) so \(y = 1\).
Find \(f'(x) = \lim _{h \rightarrow 0} \frac {f(x+h) - f(x)}{h}\) for \(f(x) = mx + b\;\) where \(m \neq 0\)

\(f'(x) = \lim _{h \rightarrow 0} m= m\).
For \(f(x) = ax^{2} + bx + c\;\) where \(a \neq 0\):
  1. find \(f'(x) = \lim _{h \rightarrow 0} \frac {f(x+h) - f(x)}{h}\) .
  2. solve \(f'(x) = 0\) for \(x\). Does this look familiar? Explain.

  1. \(f'(x) = \lim _{h \rightarrow 0} (2ax + ah + b) = 2ax + b\).
  2. Solving \(f'(x) = 2ax + b = 0\) for \(x\) gives \(x = -\frac {b}{2a}\) which is the formula for the \(x\)-coordinate of the vertex of the parabola \(y = f(x)\). If we zoom in near the the vertex of a parabola, the graph becomes locally flat so it makes sense the slope of the tangent line, \(f'(x) = 0\) there.
In Exercises diffquotexerfirstb - diffquotexerlastb, find the limit of the following difference quotients:
  1. \(\lim _{\Delta x \rightarrow 0} \frac {f(-1+\Delta x) - f(-1)}{\Delta x}\)
  2. \(\lim _{\Delta x \rightarrow 0} \frac {f(x+\Delta x) - f(x)}{\Delta x}\)
\(f(x) = \frac {2}{x}\)

  1. \(\lim _{\Delta x \rightarrow 0} \frac {2}{\Delta x-1} = -2\)
  2. \(\lim _{\Delta x \rightarrow 0} \frac {-2}{x(x+\Delta x)} = - \frac {2}{x^2}\)
\(f(x) = \frac {3}{1-x}\)

  1. \(\lim _{\Delta x \rightarrow 0} \frac {-3}{2(\Delta x - 2)} = \frac {3}{4}\)
  2. \(\lim _{\Delta x \rightarrow 0} \frac {3}{(x+\Delta x-1)(x-1)} = \frac {3}{(x-1)^2}\)
\(f(x) = \frac {1}{x^2}\)

  1. \(\lim _{\Delta x \rightarrow 0} \frac {2-\Delta x}{(\Delta x - 1)^2} =2\)
  2. \(\lim _{\Delta x \rightarrow 0}\frac {-(2x+\Delta x)}{x^2(x+\Delta x)^2} = - \frac {2}{x^3}\)
\(f(x) = \frac {2}{x+5}\)

  1. \(\lim _{\Delta x \rightarrow 0} \frac {-1}{2(\Delta x+4)} = - \frac {1}{8}\)
  2. \(\lim _{\Delta x \rightarrow 0} \frac {-2}{(x+5)(x+\Delta x+5)}= - \frac {2}{(x+5)^2}\)
In Exercises rationaltangentfirst - rationaltangentlast, find the limit of the following:
  1. \(f'(-1) = \lim _{\Delta x \rightarrow 0} \frac {f(-1+\Delta x) - f(-1)}{\Delta x}\)
  2. The equation of the tangent line at \((-1, f(-1))\).
  3. \(f'(x) = \lim _{\Delta x \rightarrow 0} \frac {f(x+\Delta x) - f(x)}{\Delta x}\)
  4. The equation of the tangent line at \((0,f(0))\).
\(f(x) = \frac {1}{4x-3}\)

  1. \(f'(-1) =\lim _{\Delta x \rightarrow 0} \frac {4}{7(4 \Delta x - 7)} = - \frac {4}{49}\)
  2. \(y = f'(-1)(x-(-1)) + f(-1) = -\frac {4}{49}(x + 1) + \left (-\frac {1}{7}\right )\) so \(y = -\frac {4}{49} \, x - \frac {11}{49}\).
  3. \(f'(x) = \lim _{\Delta x \rightarrow 0} \frac {-4}{(4x-3)(4x+4\Delta x-3)} = - \frac {4}{(4x-3)^2}\)
  4. \(y = f'(0)(x-0)+f(0) = -\frac {4}{9} (x-0) + \left (-\frac {1}{3}\right )\) so \(y = -\frac {4}{9} \, x - \frac {1}{3}\)
\(f(x) = \frac {3x}{x+2}\)

  1. \(f'(-1) = \lim _{\Delta x \rightarrow 0} \frac {6}{\Delta x + 1}= 6\)
  2. \(y = f'(-1)(x-(-1)) + f(-1) = 6(x+1) + (-3)\) so \(y = 6x+3\).
  3. \(f'(x) =\lim _{\Delta x \rightarrow 0} \frac {6}{(x+2)(x+\Delta x+2)} = \frac {6}{(x+2)^2}\)
  4. \(y = f'(0)(x-0) + f(0) = \frac {3}{2} \, (x-0)+0\) so \(y = \frac {3}{2} \, x\)
\(f(x) = \frac {x}{x - 9}\)

  1. \(f'(-1) = \lim _{\Delta x \rightarrow 0} \frac {9}{10(\Delta x - 10)} = -\frac {9}{100}\)
  2. \(y = f'(-1)(x-(-1)) + f(-1) = -\frac {9}{100} \, (x+1) + \frac {1}{10}\) so \(y = - \frac {9}{100} \, x + \frac {1}{100}\).
  3. \(f'(x) =\lim _{\Delta x \rightarrow 0} \frac {-9}{(x - 9)(x + \Delta x - 9)} = -\frac {9}{(x-9)^2}\)
  4. \(y = f'(0)(x-0) + f(0) = -\frac {1}{9} \, (x-0)+0\) so \(y = -\frac {1}{9} \, x\)
\(f(x) = \frac {x^2}{2x+1}\)

  1. \(f'(-1) = \lim _{\Delta x \rightarrow 0} \frac {\Delta x}{2 \Delta x - 1} = 0\)
  2. \(y = f'(-1)(x-(-1)) + f(-1) = (0) (x+1) + (-1)\) so \(y = -1\).
  3. \(f'(x) =\lim _{\Delta x \rightarrow 0}\frac {2x^2+2x\Delta x+2x+\Delta x}{(2x+1)(2x+2\Delta x+1)}= \frac {2x^2+2x}{(2x+1)^2}\)
  4. \(y = f'(0)(x-0) + f(0) = (0)(x-0)+0\) so \(y = 0\)
In Exercises diffquotexerfirstc - diffquotexerlastc, find the limit of the following difference quotients:
  1. \(\lim _{\Delta t \rightarrow 0} \frac {g(\Delta t) - g(0)}{\Delta t}\)
  2. \(\lim _{\Delta t \rightarrow 0} \frac {g(t+\Delta t) - g(t)}{\Delta t}\)
\(g(t) = \sqrt {9-t}\)

  1. \(\lim _{\Delta t \rightarrow 0}\frac {-1}{\sqrt {9-\Delta t} +3} = -\frac {1}{6}\)
  2. \(\lim _{\Delta t \rightarrow 0} \frac {-1}{\sqrt {9-t-\Delta t} + \sqrt {9-t}}= -\frac {1}{2 \sqrt {9-t}}\)
\(g(t) = \sqrt {2t+1}\)

  1. \(\lim _{\Delta t \rightarrow 0}\frac {2}{\sqrt {2\Delta t+1} + 1} =1\)
  2. \(\lim _{\Delta t \rightarrow 0}\frac {2}{\sqrt {2t+2\Delta t+1} + \sqrt {2t+1}}= \frac {2}{2 \sqrt {2t+1}}\)
In Exercises tangentrootfirst - tangentrootlast, find the following:
  1. \(g'(0) = \lim _{\Delta t \rightarrow 0} \frac {g(\Delta t) - g(0)}{\Delta t}\)
  2. The equation of the tangent line at \((0, g(0))\).
  3. \(g'(t) = \lim _{\Delta t \rightarrow 0} \frac {g(t+\Delta t) - g(t)}{\Delta t}\)
  4. The equation of the tangent line at \((1, g(1))\).
\(g(t) = \sqrt {-4t+5}\)

  1. \(g'(0) = \lim _{\Delta t \rightarrow 0} \frac {-4}{\sqrt {5-4\Delta t} + \sqrt {5}} = - \frac {2}{\sqrt {5}}\)
  2. \(y = g'(0) (x - 0) + g(0) = -\frac {2}{\sqrt {5}} \, (x-0) + \sqrt {5}\) so \(y = -\frac {2}{\sqrt {5}} \, x + \sqrt {5}\).
  3. \(g'(t) = \lim _{\Delta t \rightarrow 0} \frac {-4}{\sqrt {-4t-4\Delta t+5} + \sqrt {-4t+5}} = -\frac {2}{\sqrt {-4t+5}}\)
  4. \(y = g'(1)(x-1)+g(1) = (-2)(x-1) + 1\) so \(y = -2x+3\)
\(g(t) = \sqrt {4-t}\)

  1. \(g'(0) = \lim _{\Delta t \rightarrow 0} \frac {-1}{\sqrt {4-\Delta t} + 2} = - \frac {1}{4}\)
  2. \(y = g'(0) (x - 0) + g(0) = -\frac {1}{4} \, (x-0) + 2\) so \(y = -\frac {1}{4} \, x +2\).
  3. \(g'(t) = \lim _{\Delta t \rightarrow 0} \frac {-1}{\sqrt {4-t-\Delta t} + \sqrt {4-t}} = -\frac {1}{2 \sqrt {4-t}} \)
  4. \(y = g'(1)(x-1)+g(1) = -\frac {1}{2 \sqrt {3}} \, (x-1) + \sqrt {3}\) so \(y = -\frac {1}{2 \sqrt {3}} \, x + \frac {7 \sqrt {3}}{6}\).
For \(g(t) = t \sqrt {t}\):

  1. Explain why \(g'(0) = \lim _{\Delta t \rightarrow 0} \frac {g(\Delta t) - g(0)}{\Delta t}\) does not exist.

    If \(\Delta t < 0\), then \(\sqrt {\Delta t}\) is not a real number. Hence, \(g'(0) = \lim _{\Delta t \rightarrow 0} \frac {g(\Delta t) - g(0)}{\Delta t}\) does not exist.

  2. Find the derivative from the right at \(t=0\): \(g_{+}'(0) = \lim _{\Delta t \rightarrow 0^{+}} \frac {g(\Delta t) - g(0)}{\Delta t}\)

    \(g_{+}'(0) =\lim _{\Delta t \rightarrow 0^{+}} (\Delta t)^{\frac {1}{2}} = 0\)

  3. Find \(y = g_{+}'(0) (x-0) + g(0)\) and interpret.

    \(y = g_{+}'(0) (x - 0) + g(0) = (0) (x-0) + 0\) so \(y = 0\)

    This line is a tangent line to the graph of \(y = t \sqrt {t}\) at \((0,0)\) for \(t \geq 0\).

  4. Find \(g'(t) = \lim _{\Delta t \rightarrow 0} \frac {g(t+\Delta t) - g(t)}{\Delta t}\). Assume \(t>0\).

    \(g'(t) =\lim _{\Delta t \rightarrow 0}\frac {3t^2+3t\Delta t+(\Delta t)^2}{(t+\Delta t)^{3/2} + t^{3/2}} = \frac {3t^2}{2 t^{3/2}} = \frac {3}{2} \, t^{1/2}\) provided \(t > 0\).
Let \(g(t) = \sqrt {at+b}\), \(a \neq 0\).

  1. Find \(g'(t) = \lim _{\Delta t \rightarrow 0} \frac {g(t+\Delta t) - g(t)}{\Delta t}\)

    \(g'(t) = \lim _{\Delta t \rightarrow 0} \frac {a}{\sqrt {at+a\Delta t+b} + \sqrt {at+b}} = \frac {a}{2 \sqrt {at+b}}\).

  2. What restrictions do you place on \(t\) so your formula is valid?

    We are told \(a \neq 0\). In order for the square roots to be happy, we assume \(at + b > 0\). Hence, \(t > - \frac {b}{a}\) if \(a>0\) and \(t < - \frac {b}{a}\) if \(a<0\).
Let \(f(x) =|x|\).

  1. Explain why \(f\) is continuous at \(x = 0\).

    \(f\) is continuous at \(x=0\) since \(\lim _{x \rightarrow 0} f(x) = \lim _{x \rightarrow 0} |x| = 0 = |0| = f(0)\).

  2. Show \(f'(0)\) does not exist by showing \(\lim _{h \rightarrow 0^{-}} \frac {f(h) - f(0)}{h} =-1\) but \(\lim _{h \rightarrow 0^{+}} \frac {f(h) - f(0)}{h} = 1\).

    \(\lim _{h \rightarrow 0^{-}} \frac {f(h) - f(0)}{h} = \lim _{h \rightarrow 0^{-}} \frac {-h}{h} =-1\)

    \(\lim _{h \rightarrow 0^{+}} \frac {f(h) - f(0)}{h} = \lim _{h \rightarrow 0^{+}} \frac {h}{h} = 1\)

  3. Graph \(y = f(x)\) near \((0,0)\). Interpret your answer to number cornerex graphically.

    Near \(x = 0\), the graph of \(y = f(x)\) looks like a ‘\(\vee \)’ shape consisting of a line of slope \(-1\) to the left of \(x=0\) and a line with a slope of \(+1\) to the right of \(x = 0\).

  4. Find and simplify \(f'(x) = \lim _{h \rightarrow 0} \frac {|x+h| - |x|}{h}\) assuming \(x \neq 0\).

    HINT: Consider the two cases \(x > 0\) and \(x < 0\) …

    If \(x<0\), \(f'(x) = lim_{h \rightarrow 0} \frac {|x+h| - |x|}{h} = -1\).

    If \(x>0\), \(f'(x) = \lim _{h \rightarrow 0} \frac {|x+h| - |x|}{h} = 1\).

Let \(g(t) = \sqrt [3]{t}\).

  1. Explain why \(g\) is continuous at \(t = 0\).

    \(g\) is continuous at \(t=0\) since \(\lim _{t \rightarrow 0} g(t) = \lim _{t \rightarrow 0} \sqrt [3]{t} = 0 = \sqrt [3]{0} = g(0)\).

  2. Show \(g'(0)\) does not exist by showing \(\lim _{\Delta t \rightarrow 0} \frac {g(\Delta t) - g(0)}{\Delta t} = \infty \).

    \(\lim _{\Delta t \rightarrow 0} \frac {g(\Delta t) - g(0)}{\Delta t} = \lim _{\Delta t \rightarrow 0} \frac {1}{(\Delta t)^{\frac {2}{3}}} = \infty \)

  3. Graph \(y = g(t)\) near \((0,0)\). Interpret your answer to number verticaltangentex graphically.

    The graph of \(y = g(t)\) near \((0,0)\) is a vertical line. Since \(g\) is increasing through \((0,0)\), the ‘slope’ of this vertical line could be seen as \(+ \infty \).

  4. Find and simplify \(g'(t) = \lim _{\Delta t \rightarrow 0} \frac {g(t+\Delta t) - g(t)}{\Delta t}\) assuming \(t \neq 0\).

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

    \(g'(t) = \lim _{\Delta t \rightarrow 0} \frac {1}{(t+\Delta t)^{\frac {2}{3}} + (t+\Delta t)^{\frac {1}{3}} t^{\frac {1}{3}} + t^{\frac {2}{3}}} = \frac {1}{3 t^{\frac {2}{3}}}\), \(t \neq 0\).
Let \(h(x) = x^{\frac {2}{3}}\).

  1. Explain why \(h\) is continuous at \(x = 0\).

    \(\lim _{x \rightarrow 0} h(x) = \lim _{x \rightarrow 0} x^{\frac {2}{3}} = 0 = 0^{\frac {2}{3}} = h(0)\).

  2. Show \(h'(0)\) does not exist by showing \(\lim _{\Delta x \rightarrow 0^{-}} \frac {h(\Delta x) - h(0)}{\Delta x} = -\infty \) and \(\lim _{\Delta x \rightarrow 0^{+}} \frac {h(\Delta x) - h(0)}{\Delta x} = \infty \)

    \(\lim _{\Delta x \rightarrow 0^{-}} \frac {h(\Delta x) - h(0)}{\Delta x} = \lim _{\Delta x \rightarrow 0^{-}} \frac {1}{(\Delta x)^{\frac {1}{3}}} = -\infty \)

    \(\lim _{\Delta x \rightarrow 0^{+}} \frac {h(\Delta x) - h(0)}{\Delta x} = \lim _{\Delta x \rightarrow 0^{+}} \frac {1}{(\Delta x)^{\frac {1}{3}}} = \infty \)

  3. Graph \(y = h(x)\) near \((0,0)\). Interpret your answer to number cuspex graphically.

    The graph \(y = h(x)\) near \((0,0)\) shows a steeply decreasing graph as we approach \((0,0)\) from the left followed by a steeply increasing graph as we approach \((0,0)\) from the right.

  4. Find and simplify \(h'(x) = \lim _{\Delta x \rightarrow 0} \frac {h(x+\Delta x) - h(x)}{\Delta x}\) assuming \(x \neq 0\).

    HINT: \((a-b)\left (a^2+ab+b^2\right ) = a^3 - b^3\) and \(a^2 - b^2 = (a+b)(a-b)\).

    \(h'(x) = \lim _{\Delta x \rightarrow 0} \frac {h(x+\Delta x) - h(x)}{\Delta x} = \lim _{\Delta x \rightarrow 0} \frac {2x + \Delta x}{(x+\Delta x)^{\frac {4}{3}} + (x+\Delta x)^{\frac {2}{3}} \, x^{\frac {2}{3}} + x^{\frac {4}{3}}} = \frac {2}{3x^{\frac {1}{3}}}\), \(x \neq 0\).

Recall from Exercise JasonHammerExercise1 in Section QuadraticFunctions that Carl’s friend Jason participates in the Highland Games. In one event, the hammer throw, the height \(h(t)\) in feet of the hammer above the ground \(t\) seconds after Jason lets it go is modeled by the function \(h(t) = -16t^2 + 22.08t + 6\).

  1. Find and simplify a formula for the velocity of the hammer, \(v(t) = h'(t) = \lim _{\Delta t \rightarrow 0} \frac {h(t + \Delta t) - h(t)}{\Delta t}\).

    \(v(t) = \lim _{\Delta t \rightarrow 0} \frac {h(t + \Delta t) - h(t)}{\Delta t} = \lim _{\Delta t \rightarrow 0} \frac {-16(\Delta t)^2 -32 t \, \Delta t + 22.08 \Delta t}{\Delta t} = -32t + 22.08\).

  2. Find and interpret \(v(0)\).

    \(v(0) = -32(0) + 22.08 = 22.08\). This means initially (when Jason lets go of the hammer), the hammer is traveling upwards at \(22.08\) feet per second.

  3. Solve \(v(t) = 0\) and interpret.

    \(v(t) = 0\) when \(t = \frac {22.08}{32} = 0.69\). This means the (vertical) velocity zeros out \(0.69\) seconds after Jason lets go of the hammer. In this scenario, this corresponds to when the hammer reaches its peak height.

  4. Find the velocity of the hammer when it hits the ground, rounded to three decimal places.

    We first find when the hammer hits the ground by solving \(h(t) = 0\). The positive answer here is \(t \approx 1.612\) seconds. The velocity of the hammer is: \(v(1.612) = -32(1.612) + 22.08 = -29.504\). The hammer hits the ground going (approximately) \(29.504\) feet per second.
In Exercise fueleconomyexercise in Section QuadraticFunctions, the average fuel economy \(F(t)\) in miles per gallon (mpg) for passenger cars in the US \(t\) years after 1980 is modeled by \(F(t) = -0.0076t^2+0.45t + 16\), \(0 \leq t \leq 28\).

  1. Find and simplify a formula for \(F'(t) = \lim _{h \rightarrow 0} \frac {F(t + h) - F(t)}{h}\).

    \(F'(t) = \lim _{h \rightarrow 0} \frac {F(t + h) - F(t)}{h} = \lim _{h \rightarrow 0} \frac {-0.0076h^{2} -0.0152th+0.45h}{h} =-0.0152t + 0.45\).

  2. Find and interpret \(F'(0)\), \(F'(5)\) and \(F'(10)\).

    \(F'(0) = 0.45\), so fuel economy was increasing at a rate of \(0.45\) mpg per year in 1980.

    \(F'(5) = 0.374\), so fuel economy was increasing at a rate of \(0.374\) mpg per year in 1985.

    \(F'(10) = 0.298\), so fuel economy was increasing at a rate of \(0.298\) mpg per year in 1990.

  3. Interpret the trend you observe in your answers to part fueleconomytrendexercise.

    Based on the model, we have that during the years 1980 - 1990, fuel economy was increasing, but less so as the decade wore on. Technical and cost limitations could be at work here.
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.

  1. Find and interpret \(C'(75) = \lim _{h \rightarrow 0} \frac {C(75+h) - C(75)}{h}\).

    \(C'(75) = \lim _{h \rightarrow 0} \frac {C(75+h) - C(75)}{h} = \lim _{h \rightarrow 0} \frac {0.03h^{3}+2.25h^{2}+56.25h}{h} = 56.25\). This means when producing 75 systems, the cost is increasing at a rate of \(\$ 56.25\) per system.

  2. Recall in Exercise AverageCostMarginalCostExercise in Section FunctionArithmetic, we found the marginal cost, \(MC(75) = 58.53\), which means it will cost an additional \(\$ 58.53\) to produce the \(76\)th item. Compare \(C'(75)\) and \(MC(75)\).

    We see that \(C'(75)\) is numerically close to \(MC(75)\) but the former is a rate of change (measured in dollars per system) where the latter is a change (measured in dollars). Note that if we set \(h = 1\) in the difference quotient:
    \[\frac {C(75+h) - C(75)}{h} = \frac {C(75+1) - C(75)}{1} = C(76) - C(75)\]
    we see these two quantities can be used to approximate each other.
In Exercises MatchFcnDerivative1first - MatchFcnDerivative1last, match the graph of the function with a plausible graph of its derivative. Choose from Graph A, B, or C.
\(y = f(x)\):

[Picture]

Graph A:

[Picture]

Graph B.:

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Graph C.:

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\(y = g(x)\):

[Picture]

Graph A:

[Picture]

Graph B.:

[Picture]

Graph C.:

[Picture]

\(y = h(x)\):

[Picture]

Graph A:

[Picture]

Graph B.:

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Graph C.:

[Picture]

In Exercises MatchFcnDerivative2first - MatchFcnDerivative2last, match the graph of the function with a plausible graph of its derivative. Choose from Graph A, B, or C below.
\(y = f(x)\):

[Picture]

Graph A: [Picture] Graph B:

[Picture]

Graph C:

[Picture]

\(y = g(x)\):

[Picture]

Graph A: [Picture] Graph B:

[Picture]

Graph C:

[Picture]

\(y = h(x)\):

[Picture]

Graph A: [Picture] Graph B:

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Graph C:

[Picture]