Skip to main content
Query

4.9: Utungaji wa Kazi

  • Page ID
    164658
  • \( \newcommand{\vecs}[1]{\overset { \scriptstyle \rightharpoonup} {\mathbf{#1}} } \) \( \newcommand{\vecd}[1]{\overset{-\!-\!\rightharpoonup}{\vphantom{a}\smash {#1}}} \)\(\newcommand{\id}{\mathrm{id}}\) \( \newcommand{\Span}{\mathrm{span}}\) \( \newcommand{\kernel}{\mathrm{null}\,}\) \( \newcommand{\range}{\mathrm{range}\,}\) \( \newcommand{\RealPart}{\mathrm{Re}}\) \( \newcommand{\ImaginaryPart}{\mathrm{Im}}\) \( \newcommand{\Argument}{\mathrm{Arg}}\) \( \newcommand{\norm}[1]{\| #1 \|}\) \( \newcommand{\inner}[2]{\langle #1, #2 \rangle}\) \( \newcommand{\Span}{\mathrm{span}}\) \(\newcommand{\id}{\mathrm{id}}\) \( \newcommand{\Span}{\mathrm{span}}\) \( \newcommand{\kernel}{\mathrm{null}\,}\) \( \newcommand{\range}{\mathrm{range}\,}\) \( \newcommand{\RealPart}{\mathrm{Re}}\) \( \newcommand{\ImaginaryPart}{\mathrm{Im}}\) \( \newcommand{\Argument}{\mathrm{Arg}}\) \( \newcommand{\norm}[1]{\| #1 \|}\) \( \newcommand{\inner}[2]{\langle #1, #2 \rangle}\) \( \newcommand{\Span}{\mathrm{span}}\)\(\newcommand{\AA}{\unicode[.8,0]{x212B}}\)

    Picha inayofuata ni mara nyingine tena kutoka kwenye kitabu cha Biashara cha OER cha Biashara cha Calaway, Hoffman na Lippman, 2013 na kinatumiwa kwa ruhusa (Creative Commons Attribution 3.0 United States License).

    clipboard_ef44f8b2a672a0c527c65c7dcb7f25b19.png
    Kielelezo Template:index

    Nukuu\(f(g(x))\) na\(g(f(x))\) inaweza kuwa rahisi kuelewa kuliko kutumia operator utungaji. Kwa\(f(g(x))\), fikiria kuifunga mfuko. Zawadi imewekwa ndani ya sanduku (zawadi ni\(g(x)\), sanduku ni\(f(x)\)) na sasa iliyofungwa\(f(x)\), ina zawadi\(g(x)\).

    Ikiwa\(f(x) = x^2 − 2\) na\(g(x) =\sqrt{x}\), tafuta:

    1. \(f(g(x))\)na uwanja wa kazi ya composite
    2. \(g(f(x))\)na uwanja wa kazi ya composite
    Suluhisho
    1. Utungaji wa kazi,\(f(g(x))\) ni:

    \(\begin{aligned} f(g(x)) &&\text{ Function composition, }f \text{ of }g\text{ of }x \\ f(\sqrt{x}) &&\text{ Replace } g(x)\text{ with }\sqrt{x} \\ ( \sqrt{x})^2 − 2 && \text{ In the function } f(x)\text{, every }x \text{ is replaced with } g(x) =\sqrt{x} \\ x − 2 && f(g(x))\text{, answer simplified.} \end{aligned}\)

    Kikoa cha kazi ya composite kina vikwazo vya uwanja wa kazi ya ndani, pamoja na vikwazo vya kazi ya composite.

    uwanja wa kazi ya ndani,\(g(x) = \sqrt{x}\) ni kwamba\(x\) lazima nonnegative, au katika muda notation\([0, \infty )\)

    uwanja wa kazi Composite,\(x − 2\) ni namba zote halisi,\((−\infty , \infty )\)

    Kwa hiyo, uwanja wa\(f(g(x))\) sisi\([0, \infty )\).

    1. Utungaji wa kazi,\(g(f(x))\) ni:

    \(\begin{aligned} g(f(x)) &&\text{ Function composition, }g \text{ of } f \text{ of } x \\ g(x^2 − 2)&& \text{ Replace }f(x)\text{ with } x^2 − 2 \\ \sqrt{x^2 − 2} &&\text{ In the function } g(x)\text{, every }x \text{ is replaced with } f(x) = x^2−2 \\ x^2 − 2 && g(f(x))\text{, answer simplified. }\end{aligned}\)

    Kikoa cha kazi ya composite kina vikwazo vya uwanja wa kazi ya ndani, pamoja na kazi ya composite.

    Domain ya kazi ya ndani,\(f(x) = x^2 − 2\) ni namba zote halisi, au katika nukuu ya muda\((−\infty , \infty )\)

    Kikoa cha kazi ya composite,\(\sqrt{x^2} − 2\) ni kwamba wingi\(x^2 −2\) lazima usiwe na hasi, au\(x^2 −2 \geq 0\).

    Kutatua\(x^2 − 2 \geq 0\) kwa\(x\),\(x \geq 2\) na\(x \leq −2\). Katika nukuu ya muda,\((−\infty , −2] \cup [2, \infty )\)

    Kwa hiyo, uwanja wa kazi ya composite, g (f (x)) ni kikoa cha kuzuia zaidi,\((−\infty , −2] \cup [2, \infty )\).

    Ikiwa\(f(x) = \dfrac{1 }{x − 4}\) na\(g(x) = \dfrac{5 }{x} + 4\), tafuta:

    1. \(f(g(x))\)na uwanja wa kazi ya composite
    2. \(g(f(x))\)na uwanja wa kazi ya composite
    Suluhisho
    1. Utungaji wa kazi,\(f(g(x))\) ni:

    \(\begin{aligned} f(g(x)) \text{ Function composition, } f\text{ of }g \text{ of }x\\ f\left( \dfrac{5}{ x} + 4\right) && \text{ Replace }g(x)\text{ with }\dfrac{5 }{x} + 4 \\ \dfrac{1 }{\left(5 x + 4\right)− 4} && \text{ In the function } f(x)\text{, every x is replaced with } g(x) = \dfrac{5}{ x} + 4 \\ \dfrac{1 }{\dfrac{5 }{x}}&&\text{ Simplify} \\ \dfrac{x }{5} && f(g(x))\text{, answer simplified. }\end{aligned}\)

    Kikoa cha kazi ya composite kina vikwazo vya uwanja wa kazi ya ndani, pamoja na vikwazo vya kazi ya composite.

    Uwanja wa kazi ya ndani,\(g(x) = 5 x + 4\) ni maadili yote ya\(x\) vile ambayo\(x\) haipaswi kuwa 0, au kwa nukuu ya muda\((−\infty , 0) \cup (0, \infty )\)

    uwanja wa kazi Composite,\(\dfrac{x }{5}\) ni namba zote halisi,\((−\infty , \infty )\) Kwa hiyo, uwanja wa\(f(g(x))\) ni\((−\infty , 0) \cup (0, \infty )\)

    1. Utungaji wa kazi,\(g(f(x))\) ni

    \(\begin{aligned} g(f(x))&&\text{Function composition, } g \text{ of } f\text{ of }x \\ g\left( \dfrac{1 }{x −4}\right) &&\text{Replace } f(x) \text{ with }\dfrac{1}{ x − 4}\\ \dfrac{5 }{\dfrac{1 }{x − 4}} + 4 &&\text{In the function } g(x)\text{, every x is replaced with } f(x) = \dfrac{1 }{x − 4}\\ 5(x − 4) + 4 && \text{ Simplify the fraction} \\ 5x − 20 + 4 &&\text{ Simplify more}\\ 5x − 16 && g(f(x))\text{, answer simplified.} \end{aligned}\)

    Kikoa cha kazi ya composite kina vikwazo vya uwanja wa kazi ya ndani, pamoja na kazi ya composite.

    uwanja wa kazi ya ndani,\(f(x) = \dfrac{1}{ x − 4 }\) ni kwamba\(x\neq 4\), au katika nukuu ya muda\((−\infty , 4) \cup (4, \infty )\)

    Kikoa cha kazi ya composite,\(5x − 16\) ni namba zote halisi,\((−\infty , \infty )\).

    Kwa hiyo, uwanja wa kazi ya composite,\(g(f(x))\) ni kikoa cha kuzuia zaidi,\((−\infty , 4) \cup (4, \infty)\).

    Kwa kazi zilizopewa, tafuta wote\(f(g(x))\) na\(g(f(x))\), na upate uwanja wa kazi ya composite.

    1. \(f(x) = 3x^ 2 + x − 10\),\(g(x) = 1 − 20x\)
    2. \(f(x) = 3x − 2\),\(g(x) = \dfrac{1}{ 3} x + \dfrac{2 }{3}\)
    3. \(f(x) = 4x − 1\),\(g(x) = \sqrt{6 + 7x}\)
    4. \(f(x) = 5x + 2\),\(g(x) = x^2 − 14x\)
    5. \(f(x) = x^ 2 − 2x + 1\),\(g(x) = 8 − 3x ^2\)
    6. \(f(x) = x ^2 + 3\),\(g(x) = \sqrt{5 + x^2} \)