In Exercises funccompeval1first - funccompeval1last, use the given pair of functions to find the following values if they exist.

  • \((g\circ f)(0)\)
  • \((f\circ g)(-1)\)
  • \((f \circ f)(2)\)
  • \((g\circ f)(-3)\)
  • \((f\circ g)\left (\frac {1}{2}\right )\)
  • \((f \circ f)(-2)\)
\(f(x) = x^2\), \(g(t) = 2t+1\)
  • \((g\circ f)(0) = \answer {1}\)
  • \((f\circ g)(-1)= \answer {1}\)
  • \((f \circ f)(2)= \answer {16}\)
  • \((g\circ f)(-3)= \answer {19}\)
  • \((f\circ g)\left (\frac {1}{2}\right )= \answer {4}\)
  • \((f \circ f)(-2)= \answer {16}\)
\(f(x) = 4-x\), \(g(t) = 1-t^2\)

  • \((g\circ f)(0)\)

    \((g\circ f)(0) = -15\)

  • \((f\circ g)(-1)\)

    \((f\circ g)(-1) = 4\)

  • \((f \circ f)(2)\)

    \((f \circ f)(2) = 2\)

  • \((g\circ f)(-3)\)

    \((g\circ f)(-3) = -48\)

  • \((f\circ g)\left (\frac {1}{2}\right )\)

    \((f\circ g)\left (\frac {1}{2}\right ) = \frac {13}{4}\)

  • \((f \circ f)(-2)\)

    \((f \circ f)(-2) = -2\)
\(f(x) = 4-3x\), \(g(t) = |t|\)
  • \((g\circ f)(0) = \answer {4}\)
  • \((f\circ g)(-1)= \answer {1}\)
  • \((f \circ f)(2)= \answer {10}\)
  • \((g\circ f)(-3)= \answer {13}\)
  • \((f\circ g)\left (\frac {1}{2}\right )= \answer {\frac {5}{2}}\)
  • \((f \circ f)(-2)= \answer {-26}\)
\(f(x) = |x-1|\), \(g(t) = t^2-5\)

  • \((g\circ f)(0)\)

    \((g\circ f)(0) = -4\)

  • \((f\circ g)(-1)\)

    \((f\circ g)(-1) = 5\)

  • \((f \circ f)(2)\)

    \((f \circ f)(2) = 0\)

  • \((g\circ f)(-3)\)

    \((g\circ f)(-3) = 11\)

  • \((f\circ g)\left (\frac {1}{2}\right )\)

    \((f\circ g)\left (\frac {1}{2}\right ) = \frac {23}{4}\)

  • \((f \circ f)(-2)\)

    \((f \circ f)(-2) = 2\)
\(f(x) = 4x+5\), \(g(t) = \sqrt {t}\)
  • \((g\circ f)(0) = \answer {\sqrt {5}}\)

  • \((f\circ g)(-1)\)

    \((f\circ g)(-1)\) is not real
  • \((f \circ f)(2) = \answer {57}\)

  • \((g\circ f)(-3)\)

    \((g\circ f)(-3)\) is not real
  • \((f\circ g)\left (\frac {1}{2}\right ) = \answer {5+2\sqrt {2}}\)
  • \((f \circ f)(-2) = \answer {-7}\)
\(f(x) = \sqrt {3-x}\), \(g(t) = t^2+1\)

  • \((g\circ f)(0)\)

    \((g\circ f)(0) = 4\)

  • \((f\circ g)(-1)\)

    \((f\circ g)(-1) = 1\)

  • \((f \circ f)(2)\)

    \((f \circ f)(2) = \sqrt {2}\)

  • \((g\circ f)(-3)\)

    \((g\circ f)(-3) = 7\)

  • \((f\circ g)\left (\frac {1}{2}\right )\)

    \((f\circ g)\left (\frac {1}{2}\right ) = \frac {\sqrt {7}}{2}\)

  • \((f \circ f)(-2)\)

    \((f \circ f)(-2) = \sqrt {3 - \sqrt {5}}\)
\(f(x) = 6-x-x^2\), \(g(t) = t\sqrt {t+10}\)
  • \((g\circ f)(0) = \answer {24}\)
  • \((f\circ g)(-1) = \answer {0}\)
  • \((f \circ f)(2) = \answer {6}\)
  • \((g\circ f)(-3) = \answer {0}\)
  • \((f\circ g)\left (\frac {1}{2}\right ) = \answer {\frac {27-2\sqrt {42}}{8}}\)
  • \((f \circ f)(-2) = \answer {-14}\)
\(f(x) = \sqrt [3]{x+1}\), \(g(t) = 4t^2-t\)

  • \((g\circ f)(0)\)

    \((g\circ f)(0) = 3\)

  • \((f\circ g)(-1)\)

    \((f\circ g)(-1) = \sqrt [3]{6}\)

  • \((f \circ f)(2)\)

    \((f \circ f)(2) = \sqrt [3]{\sqrt [3]{3}+1}\)

  • \((g\circ f)(-3)\)

    \((g\circ f)(-3) = 4\sqrt [3]{4}+\sqrt [3]{2}\)

  • \((f\circ g)\left (\frac {1}{2}\right )\)

    \((f\circ g)\left (\frac {1}{2}\right ) = \frac {\sqrt [3]{12}}{2}\)

  • \((f \circ f)(-2)\)

    \((f \circ f)(-2) = 0\)
\(f(x) = \frac {3}{1-x}\), \(g(t) = \frac {4t}{t^2+1}\)
  • \((g\circ f)(0) = \answer {\frac {6}{5}}\)
  • \((f\circ g)(-1) = \answer {1}\)
  • \((f \circ f)(2) = \answer {\frac {3}{4}}\)
  • \((g\circ f)(-3) = \answer {\frac {48}{25}}\)
  • \((f\circ g)\left (\frac {1}{2}\right ) = \answer {-5}\)

  • \((f \circ f)(-2)\)

    \((f \circ f)(-2)\) is undefined
\(f(x) = \frac {x}{x+5}\), \(g(t) = \frac {2}{7-t^2}\)

  • \((g\circ f)(0)\)

    \((g\circ f)(0) = \frac {2}{7}\)

  • \((f\circ g)(-1)\)

    \((f\circ g)(-1) = \frac {1}{16}\)

  • \((f \circ f)(2)\)

    \((f \circ f)(2) = \frac {2}{37}\)

  • \((g\circ f)(-3)\)

    \((g\circ f)(-3) = \frac {8}{19}\)

  • \((f\circ g)\left (\frac {1}{2}\right )\)

    \((f\circ g)\left (\frac {1}{2}\right ) = \frac {8}{143}\)

  • \((f \circ f)(-2)\)

    \((f \circ f)(-2) = -\frac {2}{13}\)
\(f(x) = \frac {2x}{5-x^2}\), \(g(t) = \sqrt {4t+1}\)
  • \((g\circ f)(0) = \answer {1}\)

  • \((f\circ g)(-1)\)

    \((f\circ g)(-1)\) is not real
  • \((f \circ f)(2) = \answer {-\frac {8}{11}}\)
  • \((g\circ f)(-3) = \answer {\sqrt {7}}\)
  • \((f\circ g)\left (\frac {1}{2}\right ) = \answer {\sqrt {3}}\)
  • \((f \circ f)(-2) = \answer {\frac {8}{11}}\)
\(f(x) =\sqrt {2x+5}\), \(g(t) = \frac {10t}{t^2+1}\)

  • \((g\circ f)(0)\)

    \((g\circ f)(0) = \frac {5\sqrt {5}}{3}\)

  • \((f\circ g)(-1)\)

    \((f\circ g)(-1)\) is not real

  • \((f \circ f)(2)\)

    \((f \circ f)(2) = \sqrt {11}\)

  • \((g\circ f)(-3)\)

    \((g\circ f)(-3)\) is not real

  • \((f\circ g)\left (\frac {1}{2}\right )\)

    \((f\circ g)\left (\frac {1}{2}\right ) = \sqrt {13}\)

  • \((f \circ f)(-2)\)

    \((f \circ f)(-2) = \sqrt {7}\)
In Exercises funccompexp1first - funccompexp1last, use the given pair of functions to find and simplify expressions for the following functions and state the domain of each using interval notation.
  • \((g \circ f)(x)\)
  • \((f \circ g)(t)\)
  • \((f \circ f)(x)\)
\(f(x) = 2x+3\), \(g(t) = t^2-9\)

  • \((g \circ f)(x) = \answer {4x^2+12x}\)

    Domain:

    \((-\infty ,\infty )\)

  • \((f \circ g)(t)= \answer {2t^2-15}\)

    Domain:

    \((-\infty ,\infty )\)

  • \((f \circ f)(x)= \answer {4x+9}\)

    Domain:

    \((-\infty ,\infty )\)
\(f(x) = x^2 -x+1\), \(g(t) = 3t-5\)

  • \((g \circ f)(x)\)

    \((g \circ f)(x) = 3x^2-3x-2\)

    Domain:

    \((-\infty , \infty )\)

  • \((f \circ g)(t)\)

    \((f \circ g)(t) = 9t^2-33t+31\)

    Domain:

    \((-\infty , \infty )\)

  • \((f \circ f)(x)\)

    \((f \circ f)(x) = x^4-2x^3+2x^2-x+1\)

    Domain:

    \((-\infty , \infty )\)
\(f(x) = x^2-4\), \(g(t) = |t|\)

  • \((g \circ f)(x) = \answer {|x^2-4|}\)

    Domain:

    \((-\infty , \infty )\)

  • \((f \circ g)(t)= \answer {t^2-4}\)

    Domain:

    \((-\infty , \infty )\)

  • \((f \circ f)(x)= \answer {x^4-8x^2+12}\)

    Domain:

    \((-\infty , \infty )\)
\(f(x) = 3x-5\), \(g(t) = \sqrt {t}\)

  • \((g \circ f)(x)\)

    \((g \circ f)(x) = \sqrt {3x-5}\)

    Domain:

    \(\left [ \frac {5}{3}, \infty \right )\)

  • \((f \circ g)(t)\)

    \((f \circ g)(t) = 3\sqrt {t}-5\)

    Domain:

    \([0,\infty )\)

  • \((f \circ f)(x)\)

    \((f \circ f)(x) = 9x-20\)

    Domain:

    \((-\infty , \infty )\)
\(f(x) = |x+1|\), \(g(t) = \sqrt {t}\)

  • \((g \circ f)(x) = \answer {\sqrt {|x+1|}}\)

    Domain:

    \((-\infty , \infty )\)

  • \((f \circ g)(t)= \answer {\sqrt {t}+1}\)

    Domain:

    \([0,\infty )\)

  • \((f \circ f)(x)= \answer {|x+1|+1}\)

    Domain:

    \((-\infty , \infty )\)
\(f(x) = 3-x^2\), \(g(t) = \sqrt {t+1}\)

  • \((g \circ f)(x)\)

    \((g \circ f)(x) = \sqrt {4-x^2}\)

    Domain:

    \([-2,2]\)

  • \((f \circ g)(t)\)

    \((f \circ g)(t) =2-t\)

    Domain:

    \([-1, \infty )\)

  • \((f \circ f)(x)\)

    \((f \circ f)(x) = -x^4+6x^2-6\)

    Domain:

    \((-\infty , \infty )\)
\(f(x) = |x|\), \(g(t) = \sqrt {4-t}\)

  • \((g \circ f)(x) = \answer {\sqrt {4-|x|}}\)

    Domain:

    \([-4,4]\)

  • \((f \circ g)(t)= \answer {\sqrt {4-t}}\)

    Domain:

    \((-\infty , 4]\)

  • \((f \circ f)(x)= \answer {|x|}\)

    Domain:

    \((-\infty , \infty )\)
\(f(x) = x^2-x-1\), \(g(t) = \sqrt {t-5}\)

  • \((g \circ f)(x)\)

    \((g \circ f)(x) = \sqrt {x^2-x-6}\)

    Domain:

    \((-\infty , -2] \cup [3,\infty )\)

  • \((f \circ g)(t)\)

    \((f \circ g)(t) =t-6-\sqrt {t-5}\)

    Domain:

    \([5,\infty )\)

  • \((f \circ f)(x)\)

    \((f \circ f)(x) =x^4-2x^3-2x^2+3x+1\)

    Domain:

    \((-\infty , \infty )\)
\(f(x) = 3x-1\), \(g(t) = \frac {1}{t+3}\)

  • \((g \circ f)(x) = \answer {\frac {1}{3x+2}}\)

    Domain:

    \(\left (-\infty , -\frac {2}{3}\right ) \cup \left (-\frac {2}{3}, \infty \right )\)

  • \((f \circ g)(t)= \answer {-\frac {t}{t+3}}\)

    Domain:

    \(\left (-\infty , -3\right ) \cup \left (-3, \infty \right )\)

  • \((f \circ f)(x)= \answer {9x-4}\)

    Domain:

    \((-\infty , \infty )\)
\(f(x) = \frac {3x}{x-1}\), \(g(t) =\frac {t}{t-3}\)

  • \((g \circ f)(x)\)

    \((g \circ f)(x) =x\)

    Domain:

    \(\left (-\infty , 1\right ) \cup (1, \infty )\)

  • \((f \circ g)(t)\)

    \((f \circ g)(t) =t\)

    Domain:

    \(\left (-\infty , 3\right ) \cup (3,\infty )\)

  • \((f \circ f)(x)\)

    \((f \circ f)(x) = \frac {9x}{2x+1}\)

    Domain:

    \(\left (-\infty , -\frac {1}{2}\right ) \cup \left (-\frac {1}{2}, 1 \right ) \cup \left (1,\infty \right )\)
\(f(x) = \frac {x}{2x+1}\), \(g(t) = \frac {2t+1}{t}\)

  • \((g \circ f)(x) = \answer {\frac {4x+1}{x}}\)

    Domain:

    \((-\infty ,-\frac {1}{2}) \cup (-\frac {1}{2},0) \cup (0,\infty )\)

  • \((f \circ g)(t) = \answer {\frac {2t+1}{5t+2}}\)

    Domain:

    \((-\infty ,-\frac {2}{5}) \cup (-\frac {2}{5},0) \cup (0,\infty )\)

  • \((f \circ f)(x) = \answer {\frac {x}{4x+1}}\)

    Domain:

    \((-\infty ,-\frac {1}{2}) \cup (-\frac {1}{2},-\frac {1}{4}) \cup (-\frac {1}{4},\infty )\)
\(f(x) = \frac {2x}{x^2-4}\), \(g(t) =\sqrt {1-t}\)

  • \((g \circ f)(x)\)

    \((g \circ f)(x) =\sqrt {\frac {x^2-2x-4}{x^2-4}}\)

    Domain:

    \(\left (-\infty , -2\right ) \cup \left [1-\sqrt {5}, 2\right ) \cup \left [1+\sqrt {5}, \infty \right )\)

  • \((f \circ g)(t)\)

    \((f \circ g)(t) = -\frac {2\sqrt {1-t}}{t+3}\)

    Domain:

    \(\left (-\infty , -3\right ) \cup \left (-3, 1\right ]\)

  • \((f \circ f)(x)\)

    \((f \circ f)(x) = \frac {4x-x^3}{x^4-9x^2+16}\)

    Domain:

    \(\left (-\infty , -\frac {1+\sqrt {17}}{2}\right ) \cup \left (-\frac {1+\sqrt {17}}{2}, -2\right ) \cup \left (-2, \frac {1-\sqrt {17}}{2}\right ) \cup \left (\frac {1-\sqrt {17}}{2}, \frac {-1+\sqrt {17}}{2}\right ) \cup \left (\frac {-1+\sqrt {17}}{2}, 2\right ) \cup \left (2, \frac {1+\sqrt {17}}{2} \right ) \cup \left (\frac {1+\sqrt {17}}{2}, \infty \right )\)
In Exercises threefunccompfirst - threefunccomplast, use \(f(x) = -2x\), \(g(t) = \sqrt {t}\) and \(h(s) = |s|\) to find and simplify expressions for the following functions and state the domain of each using interval notation.
\((h\circ g \circ f)(x)\)

\((h\circ g \circ f)(x)= |\sqrt {-2x}|= \sqrt {-2x}\), domain: \((-\infty , 0]\)
\((h\circ f \circ g)(t)\)

\((h\circ f \circ g)(t) = |-2\sqrt {t}|= 2\sqrt {t}\), domain: \([0,\infty )\)
\((g\circ f \circ h)(s)\)

\((g\circ f \circ h)(s) = \sqrt {-2|s|}\), domain: \(\{0\}\)
\((g\circ h \circ f)(x)\)

\((g\circ h \circ f)(x) = \sqrt {|-2x|} = \sqrt {2|x|}\), domain: \((-\infty , \infty )\)
\((f\circ h \circ g)(t)\)

\((f\circ h \circ g)(t) = -2|\sqrt {t}| = -2\sqrt {t}\), domain: \([0,\infty )\)
\((f\circ g \circ h)(s)\)

\((f\circ g \circ h)(s) = -2\sqrt {|s|}\), , domain: \((-\infty ,\infty )\)
In Exercises pointcompexfirst - pointcompexlast, let \(f\) be the function defined by
\[f = \{(-3, 4), (-2, 2), (-1, 0), (0, 1), (1, 3), (2, 4), (3, -1)\}\]
and let \(g\) be the function defined by
\[g = \{(-3, -2), (-2, 0), (-1, -4), (0, 0), (1, -3), (2, 1), (3, 2)\}.\]
Find the following, if it exists.
\((f \circ g)(3)\)

\((f \circ g)(3)= f(g(3)) = f(2) = 4\)
\(f(g(-1))\)

\(f(g(-1)) = f(-4)\) which is undefined
\((f \circ f)(0)\)

\((f \circ f)(0) = f(f(0)) = f(1) = 3\)
\((f \circ g)(-3)\)

\((f \circ g)(-3) = f(g(-3)) = f(-2) = 2\)
\((g \circ f)(3)\)

\((g \circ f)(3) = g(f(3)) = g(-1) = -4\)
\(g(f(-3))\)

\(g(f(-3)) = g(4)\) which is undefined
\((g \circ g)(-2)\)

\((g \circ g)(-2) = g(g(-2)) = g(0) = 0\)
\((g \circ f)(-2)\)

\((g \circ f)(-2) = g(f(-2)) = g(2) = 1\)
\(g(f(g(0)))\)

\(g(f(g(0))) = g(f(0)) = g(1) = -3\)
\(f(f(f(-1)))\)

\(f(f(f(-1))) = f(f(0)) = f(1) = 3\)
\(f(f(f(f(f(1)))))\)

\(f(f(f(f(f(1))))) = f(f(f(f(3)))) =\\ f(f(f(-1))) = f(f(0)) = f(1) = 3\)
\(\underbrace {(g \circ g \circ \cdots \circ g)}_{\mbox {$n$ times}}(0)\)

\(\underbrace {(g \circ g \circ \cdots \circ g)}_{\mbox {$n$ times}}(0) = 0\)
Find the domain and range of \(f \circ g\) and \(g \circ f\).

  • The domain of \(f \circ g\) is \(\{ -3, -2, 0, 1, 2, 3\}\) and the range of \(f \circ g\) is \(\{1, 2, 3, 4\}\).
  • The domain of \(g \circ f\) is \(\{ -2, -1, 0, 1, 3 \}\) and the range of \(g \circ f\) is \(\{ -4, -3, 0, 1, 2 \}\).

In Exercises twofuncgraphcompfirst - twofuncgraphcomplast, use the graphs of \(y=f(x)\) and \(y=g(x)\) below to find the following if it exists.

\((g\circ f)(1)\)

\((g\circ f)(1) = 3\)
\((f \circ g)(3)\)

\((f \circ g)(3) = 1\)
\((g\circ f)(2)\)

\((g\circ f)(2) = 0\)
\((f\circ g)(0)\)

\((f\circ g)(0) = 1\)
\((f\circ f)(4)\)

\((f\circ f)(4) = 1\)
\((g \circ g)(1)\)

\((g \circ g)(1) = 0\)
Find the domain and range of \(f \circ g\) and \(g \circ f\).

  • The domain of \(f \circ g\) is \([0,3]\) and the range of \(f \circ g\) is \([1, 4.5]\).
  • The domain of \(g \circ f\) is \([0,2] \cup [3,4]\) and the range is \([0,3]\).
In Exercises breakdowncompexfirst - breakdownxomexlast, write the given function as a composition of two or more non-identity functions. (There are several correct answers, so check your answer using function composition.)
\(p(x) = (2x+3)^3\)

Let \(f(x) = 2x+3\) and \(g(x) = x^3\), then \(p(x) = (g\circ f)(x)\).
\(P(x) = \left (x^2-x+1\right )^5\)

Let \(f(x) = x^2-x+1\) and \(g(x) = x^5\), \(P(x) =(g\circ f)(x)\).
\(h(t) = \sqrt {2t-1}\)

Let \(f(t) = 2t-1\) and \(g(t) = \sqrt {t}\), then \(h(t) = (g\circ f)(t)\).
\(H(t) = |7-3t|\)

Let \(f(t) = 7-3t\) and \(g(t) = |t|\), then \(H(t) = (g\circ f)(t)\).
\(r(s) = \frac {2}{5s+1}\)

Let \(f(s) = 5s+1\) and \(g(s) = \frac {2}{s}\), then \(r(s) =(g\circ f)(s)\).
\(R(s) = \frac {7}{s^2-1}\)

Let \(f(s) = s^2-1\) and \(g(s) = \frac {7}{s}\), then \(R(s) =(g\circ f)(s)\).
\(q(z) = \frac {|z|+1}{|z|-1}\)

Let \(f(z) = |z|\) and \(g(z) = \frac {z+1}{z-1}\), then \(q(z) =(g\circ f)(z)\).
\(Q(z) = \frac {2z^3+1}{z^3-1}\)

Let \(f(z) = z^3\) and \(g(z)= \frac {2z+1}{z-1}\), then \(Q(z) =(g\circ f)(z)\).
\(v(x) = \frac {2x+1}{3-4x}\)

Let \(f(x) =2x\) and \(g(x) = \frac {x+1}{3-2x}\), then \(v(x) =(g\circ f)(x)\).
\(w(x) = \frac {x^2}{x^4+1}\)

Let \(f(x) = x^2\) and \(g(x) = \frac {x}{x^2+1}\), then \(w(x) =(g\circ f)(x)\).
Write the function \(F(x) = \sqrt {\frac {x^{3} + 6}{x^{3} - 9}}\) as a composition of three or more non-identity functions.

\(F(x) = \sqrt {\frac {x^{3} + 6}{x^{3} - 9}} = (h(g(f(x)))\) where \(f(x) = x^{3}, \, g(x) = \frac {x + 6}{x - 9}\) and \(h(x) = \sqrt {x}\).
Let \(g(x) = -x, \, h(x) = x + 2, \, j(x) = 3x\) and \(k(x) = x - 4\). In what order must these functions be composed with \(f(x) = \sqrt {x}\) to create \(F(x) = 3\sqrt {-x + 2} - 4\)?

\(F(x) = 3\sqrt {-x + 2} - 4 = k(j(f(h(g(x)))))\)
What linear functions could be used to transform \(f(x) = x^{3}\) into \(F(x) = -\frac {1}{2}(2x - 7)^{3} + 1\)? What is the proper order of composition?

One solution is \(F(x) = -\frac {1}{2}(2x - 7)^{3} + 1 = k(j(f(h(g(x)))))\) where \(g(x) = 2x, \, h(x) = x - 7, \, j(x) = -\frac {1}{2}x\) and \(k(x) = x + 1\). You could also have \(F(x) = H(f(G(x)))\) where \(G(x) = 2x - 7\) and \(H(x) = -\frac {1}{2}x + 1\).
Let \(f(x) = 3x+1\) and let \(g(x) = \begin{cases} 2x-1 & \text {if $x \leq 3$} \\ 4-x & \text {if $x > 3$} \\ \end{cases}\). Find expressions for \((f \circ g)(x)\) and \((g \circ f)(x)\).

\((f \circ g)(x) = \begin{mycases} 6x-2 & \text {if $x \leq 3$} \\ 13-3x & \text {if $x > 3$} \\ \end{mycases}\) and \((g \circ f)(x) = \begin{mycases} 6x+1 & \text {if $x \leq \frac {2}{3}$} \\ 3-3x & \text {if $x > \frac {2}{3}$} \\ \end{mycases}\)
The volume \(V\) of a cube is a function of its side length \(x\). Let’s assume that \(x = t + 1\) is also a function of time \(t\), where \(x\) is measured in inches and \(t\) is measured in minutes. Find a formula for \(V\) as a function of \(t\).

\(V(x) = x^{3}\) so \(V(x(t)) = (t + 1)^{3}\)
Suppose a local vendor charges \(\$2\) per hot dog and that the number of hot dogs sold per hour \(x\) is given by \(x(t) = -4t^2+20t+92\), where \(t\) is the number of hours since \(10\) AM, \(0 \leq t \leq 4\).

  1. Find an expression for the revenue per hour \(R\) as a function of \(x\).

    \(R(x) = 2x\)

  2. Find and simplify \(\left (R \circ x\right )(t)\). What does this represent?

    \(\left (R \circ x \right )(t) = -8t^2+40t+184\), \(0 \leq t \leq 4\). This gives the revenue per hour as a function of time.

  3. What is the revenue per hour at noon?

    Noon corresponds to \(t=2\), so \(\left (R \circ x \right )(2) = 232\). The hourly revenue at noon is \(\$232\) per hour.

  4. Using Example surfaceareaex2chainrule as a guide, verify \(\frac {\Delta [R(x)]}{\Delta x} \cdot \frac {\Delta [x(t)]}{\Delta t} = \frac {\Delta [R(t)]}{\Delta t}\).

    \(\frac {\Delta [R(x)]}{\Delta x} = 2\), \(\frac {\Delta [x(t)]}{\Delta t} = -8t + 4 \Delta t + 20\).

    \(\frac {\Delta [R(x)]}{\Delta x} \cdot \frac {\Delta [x(t)]}{\Delta t} = (2)(-8t + 4 \Delta t + 20) = -16t + 8 \Delta t + 40 = \frac {\Delta [R(t)]}{\Delta t} \, \checkmark \)

The book in Example dragforceex plunges into a lake and generates a circular wave pattern. If the waves are tracked as traveling at a constant \(0.5\) meters per second \(\left ( \frac {\text {m}}{\text {s}}\right )\), use Theorem relatedratesaroc to find the rate at which the area of the disturbance is changing with respect to time as the radius changes from \(r = 1\) to \(r = 1.1\) meters (m). Be sure to include units on your answer.

Recall the area, \(A\), enclosed by a circle of radius \(r\) is given by \(A = \pi \, r^2\). Here, \(\frac {\Delta r}{\Delta t} = 0.5 \, \frac {\text {m}}{\text {s}}\).

\(\frac {\Delta A}{\Delta t} = \frac {\Delta A}{\Delta r} \cdot \frac {\Delta r}{\Delta t}\). \(\frac {\Delta A}{\Delta r} = \frac {A(1.1) - A(1)}{1.1-1} = \frac {\pi (1.1)^2 - \pi (1)^2}{0.1} = 2.1 \, \frac {\text {m}^2}{\text {m}}\), \(\frac {\Delta r}{\Delta t} = 0.5 \, \frac {\text {m}}{\text {s}}\).

Hence, \(\frac {\Delta A}{\Delta t} = \left ( 2.1 \, \frac {\text {m}^2}{\text {m}} \right ) \left (0.5 \, \frac {\text {m}}{\text {s}} \right ) = 1.05 \, \frac {\text {m}^2}{\text {s}}\)

Perfectly fine precalculus textbooks which have no Calculus content are being fed into a shredder at a rate of 2 books per minute in order to make room for precalculus textbooks with Calculus content. The shredder creates a pile of debris which is in the shape of a right circular cone whose height is twice its width.

  1. Assume the volume of the conical pile, \(V\), is given by \(V = \frac {1}{3} \, \pi r^2 h\) where \(r\) is the radius of the base of the pile and \(h\) is the height of the pile. Given the pile is twice as tall as it is wide, show we can write \(V = \frac {4}{3} \, \pi r^3\).

    The ‘width’ of the pile is the diameter of the circular base of the pile. Since the diameter of a circle is twice the radius, \(h = 2 (2r) = 4r\). Hence, \(V = \frac {1}{3} \, \pi r^2 h = \frac {1}{3} \, \pi r^2 (4r) = \frac {4}{3} \, \pi r^3\).

  2. Assuming a typical precalculus textbook is \(0.10\) cubic feet \(\left ( \text {ft}^3 \right )\), use Theorem relatedratesaroc to find the rate of change of the radius of the pile with respect to time as the radius changes from \(2\) to \(2.1\) feet. Be sure to include units on your answer.

    We have \(\frac {\Delta V}{\Delta t} = \frac {\Delta V}{ \Delta r} \cdot \frac {\Delta r}{\Delta t}\). Two textbooks per minute into the shredder amounts to the volume of the cone increasing at a rate of \(2(0.1) = 0.2 \, \frac {\text {ft}^3}{\text {min}}\). \(\frac {\Delta V}{\Delta r}= \frac {V(2.1) - V(2)}{2.1 - 2} = 16.81\overline {3} \, \pi \, \frac {\text {ft}^3}{\text {ft}}\).

    Hence, \(0.2 \, \frac {\text {ft}^3}{\text {min}} = 16.81\overline {3} \, \pi \, \frac {\text {ft}^3}{\text {ft}} \, \frac {\Delta r}{\Delta t}\) so \(\frac {\Delta r}{\Delta t} = \frac {0.2}{ 16.81\overline {3} \, \pi } \approx 0.0038 \, \frac {\text {ft}}{\text {min}}\).

Discuss with your classmates how ‘real-world’ processes such as filling out federal income tax forms or computing your final course grade could be viewed as a use of function composition. Find a process for which composition with itself (iteration) makes sense.