In Exercises basicarithonefirst - basicarithonelast , use the pair of functions \(f\) and \(g\) to find each of the following if they exist.
\((f+g)(2)\)
\((f-g)(-1)\)
\((g-f)(1)\)
\((fg)\left (\frac {1}{2}\right )\)
\(\left (\frac {f}{g}\right )(0)\)
\(\left (\frac {g}{f}\right )\left (-2\right )\)
\(f(x) = 3x+1\) and
\(g(t) = 4-t\)
\((f+g)(2)=\answer {9}\)
\((f-g)(-1)=\answer {-7}\)
\((g-f)(1)=\answer {-1}\)
\((fg)\left (\frac {1}{2}\right )=\answer {\frac {35}{4}}\)
\(\left (\frac {f}{g}\right )(0)=\answer {\frac {1}{4}}\)
\(\left (\frac {g}{f}\right )\left (-2\right )=\answer {-\frac {6}{5}}\)
\(f(x) = x^2\) and
\(g(t) = -2t+1\)
\((f+g)(2)=\answer {1}\)
\((f-g)(-1)=\answer {-2}\)
\((g-f)(1)=\answer {-2}\)
\((fg)\left (\frac {1}{2}\right )=\answer {0}\)
\(\left (\frac {f}{g}\right )(0)=\answer {0}\)
\(\left (\frac {g}{f}\right )\left (-2\right )=\answer {\frac {5}{4}}\)
\(f(x) = x^2 - x\) and
\(g(t) = 12-t^2\)
\((f+g)(2)=\answer {10}\)
\((f-g)(-1)=\answer {-9}\)
\((g-f)(1)=\answer {11}\)
\((fg)\left (\frac {1}{2}\right )=\answer {-\frac {47}{16}}\)
\(\left (\frac {f}{g}\right )(0)=\answer {0}\)
\(\left (\frac {g}{f}\right )\left (-2\right )=\answer {\frac {4}{3}}\)
\(f(x) = 2x^3\) and
\(g(t) = -t^2-2t-3\)
\((f+g)(2)=\answer {5}\)
\((f-g)(-1)=\answer {0}\)
\((g-f)(1)=\answer {-8}\)
\((fg)\left (\frac {1}{2}\right )=\answer {-\frac {17}{16}}\)
\(\left (\frac {f}{g}\right )(0)=\answer {0}\)
\(\left (\frac {g}{f}\right )\left (-2\right )=\answer {\frac {3}{16}}\)
\(f(x) = \sqrt {x+3}\) and
\(g(t) = 2t-1\)
\((f+g)(2)=\answer {3+\sqrt {5}}\)
\((f-g)(-1)=\answer {3+\sqrt {2}}\)
\((g-f)(1)=\answer {-1}\)
\((fg)\left (\frac {1}{2}\right )=\answer {0}\)
\(\left (\frac {f}{g}\right )(0)=\answer {-\sqrt {3}}\)
\(\left (\frac {g}{f}\right )\left (-2\right )=\answer {-5}\)
\(f(x) = \sqrt {4-x}\) and
\(g(t) = \sqrt {t+2}\)
\((f+g)(2)=\answer {2+\sqrt {2}}\)
\((f-g)(-1)=\answer {-1+\sqrt {5}}\)
\((g-f)(1)=\answer {0}\)
\((fg)\left (\frac {1}{2}\right )=\answer {\frac {\sqrt {35}}{2}}\)
\(\left (\frac {f}{g}\right )(0)=\answer {\sqrt {2}}\)
\(\left (\frac {g}{f}\right )\left (-2\right )=\answer {0}\)
\(f(x) = 2x\) and
\(g(t) = \dfrac {1}{2t+1}\)
\((f+g)(2)=\answer {\frac {21}{5}}\)
\((f-g)(-1)=\answer {-1}\)
\((g-f)(1)=\answer {-\frac {5}{3}}\)
\((fg)\left (\frac {1}{2}\right )=\answer {\frac {1}{2}}\)
\(\left (\frac {f}{g}\right )(0)=\answer {0}\)
\(\left (\frac {g}{f}\right )\left (-2\right )=\answer {\frac {1}{12}}\)
\(f(x) = x^2\) and
\(g(t) = \dfrac {3}{2t-3}\)
\((f+g)(2)=\answer {7}\)
\((f-g)(-1)=\answer {\frac {8}{5}}\)
\((g-f)(1)=\answer {-4}\)
\((fg)\left (\frac {1}{2}\right )=\answer {-\frac {3}{8}}\)
\(\left (\frac {f}{g}\right )(0)=\answer {0}\)
\(\left (\frac {g}{f}\right )\left (-2\right )=\answer {-\frac {3}{28}}\)
\(f(x) = x^2\) and
\(g(t) = \dfrac {1}{t^2}\)
\(f(x) = x^2+1\) and
\(g(t) = \dfrac {1}{t^2+1}\)
\((f+g)(2)=\answer {\frac {26}{5}}\)
\((f-g)(-1)=\answer {\frac {3}{2}}\)
\((g-f)(1)=\answer {-\frac {3}{2}}\)
\((fg)\left (\frac {1}{2}\right )=\answer {1}\)
\(\left (\frac {f}{g}\right )(0)=\answer {1}\)
\(\left (\frac {g}{f}\right )\left (-2\right )=\answer {\frac {1}{25}}\)
Exercises
arithfromgraphfirst -
arithfromgraphlast refer to the functions
\(f\) and
\(g\) whose graphs are below.
\((f + g)(-4)=\answer {-5}\)
\((f + g)(0)=\answer {5}\)
\((f- g)(4)=\answer {-5}\)
\((fg)(-4)=\answer {6}\)
\((fg)(-2)=\answer {0}\)
\((fg)(4)=\answer {-6}\)
\(\left (\dfrac {f}{g}\right )(0)=\answer {\dfrac {3}{2}}\)
\(\left (\dfrac {f}{g}\right )(2)=\answer {0}\)
\(\left (\dfrac {g}{f}\right )(-1)=\answer {0}\)
Find the domains of
\(f+g\) ,
\(f-g\) ,
\(fg\) ,
\(\dfrac {f}{g}\) and
\(\dfrac {g}{f}\) .
The domains of \(f+g\) , \(f-g\) and \(fg\) are all \([-4,4]\) . The domain of \(\frac {f}{g}\) is \([-4, -1) \cup (-1,4]\) and the domain of \(\frac {g}{f}\) is \([-4, -2) \cup (-2,2) \cup (2, 4]\) .
In Exercises
reformarithfirst -
reformarithlast , 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
\[g = \{(-3, -2), (-2, 0), (-1, -4), (0, 0), (1, -3), (2, 1), (3, 2)\}\]
Compute the indicated value if it
exists.
\((f + g)(-3)=\answer {2}\)
\((f - g)(2)=\answer {3}\)
\((fg)(-1)=\answer {0}\)
\((g + f)(1)=\answer {0}\)
\((g - f)(3)=\answer {3}\)
\((gf)(-3)=\answer {-8}\)
\(\left (\frac {f}{g}\right )(-2)\)
\(\left (\frac {f}{g}\right )(-1)=\answer {0}\)
\(\left (\frac {f}{g}\right )(2)=\answer {4}\)
\(\left (\frac {g}{f}\right )(-1)\)
\(\left (\frac {g}{f}\right )(3)=\answer {-2}\)
\(\left (\frac {g}{f}\right )(-3)=\answer {-\frac {1}{2}}\)
In Exercises
basicarithtwofirst -
basicarithtwolast , use the pair of functions
\(f\) and
\(g\) to find the domain of the indicated function then find and simplify an
expression for it.
\((f+g)(x)\)
\((f-g)(x)\)
\((fg)(x)\)
\(\left (\frac {f}{g}\right )(x)\)
\(f(x) = 2x+1\) and
\(g(x) = x-2\)
\((f+g)(x)=\answer {3x-1}\)
Domain: \((-\infty , \infty )\)
\((f-g)(x)=\answer {x+3}\)
Domain: \((-\infty , \infty )\)
\((fg)(x)=\answer {2x^2-3x-2}\)
Domain: \((-\infty , \infty )\)
\(\left (\frac {f}{g}\right )(x)=\answer {\frac {2x+1}{x-2}}\)
Domain: \((-\infty , 2) \cup (2, \infty )\)
\(f(x) = 1-4x\) and
\(g(x) = 2x-1\)
\((f+g)(x)=\answer {-2x}\)
Domain: \((-\infty , \infty )\)
\((f-g)(x)=\answer {2-6x}\)
Domain: \((-\infty , \infty )\)
\((fg)(x)=\answer {-8x^2+6x-1}\)
Domain: \((-\infty , \infty )\)
\(\left (\frac {f}{g}\right )(x)=\answer { \frac {1-4x}{2x-1}}\)
Domain: \(\left (-\infty , \frac {1}{2} \right ) \cup \left (\frac {1}{2}, \infty \right )\)
\(f(x) = x^2\) and
\(g(x) = 3x-1\)
\((f+g)(x)=\answer {x^2+3x-1}\)
Domain: \((-\infty , \infty )\)
\((f-g)(x)=\answer {x^2-3x+1}\)
Domain: \((-\infty , \infty )\)
\((fg)(x)=\answer {3x^3-x^2}\)
Domain: \((-\infty , \infty )\)
\(\left (\frac {f}{g}\right )(x)=\answer {\frac {x^2}{3x-1}}\)
Domain: \(\left (-\infty , \frac {1}{3} \right ) \cup \left (\frac {1}{3}, \infty \right )\)
\(f(x) = x^2-x\) and
\(g(x) = 7x\)
\((f+g)(x)=\answer {x^2+6x}\)
Domain: \((-\infty , \infty )\)
\((f-g)(x)=\answer {x^2-8x}\)
Domain: \((-\infty , \infty )\)
\((fg)(x)=\answer {7x^3-7x^2}\)
Domain: \((-\infty , \infty )\)
\(\left (\frac {f}{g}\right )(x)=\answer {\frac {x-1}{7}}\)
Domain: \(\left (-\infty , 0 \right ) \cup \left (0, \infty \right )\)
\(f(x) = x^2-4\) and
\(g(x) = 3x+6\)
\((f+g)(x)=\answer {x^2+3x+}\) Domain: \((-\infty , \infty )\)
\((f-g)(x)=\answer {x^2-3x-10}\)
Domain: \((-\infty , \infty )\)
\((fg)(x)=\answer {3x^3+6x^2-12x-24}\)
Domain: \((-\infty , \infty )\)
\(\left (\frac {f}{g}\right )(x)=\answer {\frac {x-2}{3}}\)
Domain: \(\left (-\infty , -2 \right ) \cup \left (-2, \infty \right )\)
\(f(x) = -x^2+x+6\) and
\(g(x) = x^2-9\)
\((f+g)(x)=\answer {x-3}\)
Domain: \((-\infty , \infty )\)
\((f-g)(x)=\answer {-2x^2+x+15}\)
Domain: \((-\infty , \infty )\)
\((fg)(x)=\answer {-x^4+x^3+15x^2-9x-54}\)
Domain: \((-\infty , \infty )\)
\(\left (\frac {f}{g}\right )(x)=\answer {-\frac {x+2}{x+3}}\)
Domain: \(\left (-\infty , -3 \right ) \cup \left (-3, 3 \right ) \cup (3, \infty )\)
\(f(x) = \dfrac {x}{2}\) and
\(g(x) = \dfrac {2}{x}\)
\((f+g)(x)=\answer {\frac {x^2+4}{2x}}\)
Domain: \((-\infty , 0) \cup (0, \infty )\)
\((f-g)(x)=\answer {\frac {x^2-4}{2x}}\)
Domain: \((-\infty ,0) \cup (0, \infty )\)
\((fg)(x)=\answer {1}\)
Domain: \((-\infty ,0) \cup (0, \infty )\)
\(\left (\frac {f}{g}\right )(x)=\answer {\frac {x^2}{4}}\)
Domain: \((-\infty ,0) \cup (0, \infty )\)
\(f(x) =x-1\) and
\(g(x) = \dfrac {1}{x-1}\)
\((f+g)(x)=\answer {\frac {x^2-2x+2}{x-1}}\)
Domain: \((-\infty , 1) \cup (1, \infty )\)
\((f-g)(x)=\answer {\frac {x^2-2x}{x-1}}\)
Domain: \((-\infty ,1) \cup (1, \infty )\)
\((fg)(x)=\answer {1}\)
Domain: \((-\infty ,1) \cup (1, \infty )\)
\(\left (\frac {f}{g}\right )(x)=\answer {x^2-2x+1}\)
Domain: \((-\infty ,1) \cup (1, \infty )\)
\(f(x) = x\) and
\(g(x) = \sqrt {x+1}\)
\((f+g)(x)=\answer {x+\sqrt {x+1}}\)
\((f-g)(x)=\answer {x-\sqrt {x+1}}\)
\((fg)(x)=\answer {x\sqrt {x+1}}\)
\(\left (\frac {f}{g}\right )(x)=\answer {\frac {x}{\sqrt {x+1}}}\)
\(f(x) =\sqrt {x-5}\) and
\(g(x) = f(x) = \sqrt {x-5}\)
In Exercises
decomposebasicfirst -
decomposebasiclast , write the given function as a nontrivial decomposition of functions as directed.
For
\(p(z) = 4z-z^3\) , find functions
\(f\) and
\(g\) so that
\(p=f-g\) .
One solution is \(f(z) = 4z\) and \(g(z) = z^3\) .
For
\(p(z) = 4z-z^3\) , find functions
\(f\) and
\(g\) so that
\(p=f+g\) .
One solution is \(f(z) = 4z\) and \(g(z) = - z^3\) .
For
\(g(t) = 3t|2t-1|\) , find functions
\(f\) and
\(h\) so that
\(g = fh\) .
One solution is \(f(t) = 3t\) and \(h(t) = |2t-1|\)
For
\(r(x) = \dfrac {3-x}{x+1}\) , find functions
\(f\) and
\(g\) so
\(r = \dfrac {f}{g}\) .
One solution is \(f(x) = 3-x\) and \(g(x) = x+1\) .
For
\(r(x) = \dfrac {3-x}{x+1}\) , find functions
\(f\) and
\(g\) so
\(r = fg\) .
One solution is \(f(x) = 3-x\) and \(g(x) = (x+1)^{-1}\) .
Can
\(f(x) = x\) be decomposed as
\(f = g-h\) where
\(g(x) = x+\dfrac {1}{x}\) and
\(h(x) = \dfrac {1}{x}\) ?
No. The equivalence does not hold when \(x = 0\) .
In this exercise, we explore decomposing a function into its positive and negative parts. Given a function
\(f\) , we define the
positive part of
\(f\) , denoted
\(f_{+}\) and
negative part of
\(f\) , denoted
\(f_{-}\) by:
\[ f_{+}(x) = \dfrac {f(x) + |f(x)|}{2}, \qquad \text {and} \qquad f_{-}(x) = \dfrac {f(x) - |f(x)|}{2}. \]
Using a graphing utility, graph each of the functions \(f\) below along with \(f_{+}\) and \(f_{-}\) .
\(f(x) = x-3\)
\(f(x) = x^2-x-6\)
\(f(x) = 4x-x^3\)
Why is \(f_{+}\) called the ‘positive part’ of \(f\) and \(f_{-}\) called the ‘negative part’ of \(f\) ?
Show that \(f = f_{+} + f_{-}\) .
Use Definition absolutevaluepiecewise to rewrite the expressions for \(f_{+}(x)\) and \(f_{-}(x)\) as piecewise defined functions.
![[Picture]](ExercisesforFunctionArithmetic.online-08b2af58dc79b05320285aec9a8541d9.svg)