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Yuav ua li cas koj multiply composite functions?
Yuav ua li cas koj multiply composite functions?

Video: Yuav ua li cas koj multiply composite functions?

Video: Yuav ua li cas koj multiply composite functions?
Video: Cas yim laus es pheej yim nco.9/4/2018 2024, Kaum ib hlis
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Multiplication thiab Composition of Functions

  1. Rau muab ntau ntau ib muaj nuj nqi los ntawm scalar, muab ntau ntau txhua qhov tso zis los ntawm qhov scalar.
  2. Thaum peb coj f (g(x)), peb muab g(x) raws li cov tswv yim ntawm tus muaj nuj nqi f.
  3. Piv txwv li, yog f (x) = 10x thiab g(x) = x + 1, ces nrhiav f (g(4)), peb nrhiav g(4) = 4 + 1 + 5, thiab ces ntsuas f (5) Luas = 10 (5) = 50.
  4. Piv txwv li: f(x) = 2x - 2, g(x) = x2 - 8.

Raws li, koj yuav ua li cas ntau txoj haujlwm?

Multiplication ntawm Muaj nuj nqi Rau muab ntau ntau ib muaj nuj nqi los ntawm lwm tus muaj nuj nqi , muab ntau ntau lawv outputs. Piv txwv li, yog f (x) = 2x thiab g(x) = x + 1, ces fg(3) = f (3) × g(3) = 6 × 4 = 24. fg(x) = 2x(x) + 1) x = 22 + x.

Tsis tas li ntawd, koj ua haujlwm li cas? Xav txog qhov muaj nuj nqi f(x) = 2 x + 1. Peb lees paub qhov sib npaug y = 2 x + 1 raws li daim ntawv Slope-Intercept ntawm kab zauv ntawm txoj kab nqes 2 thiab y-intercept (0, 1). Xav txog qhov taw tes txav mus rau qhov daim duab ntawm f. Raws li qhov taw tes txav mus rau sab xis nws nce.

Tom qab ntawd, ib tug kuj yuav nug, dab tsi yog cov khoom ntawm ob lub zog?

Thaum koj muab ntau ob lub luag haujlwm ua ke, koj yuav tau txais peb muaj nuj nqi raws li qhov tshwm sim, thiab qhov thib peb muaj nuj nqi yuav yog khoom ntawm ob tus thawj muaj nuj nqi . Piv txwv li, yog tias koj muab f(x) thiab g(x), lawv khoom yuav h(x)=fg(x), or h(x)=f(x)g(x). Koj tseem tuaj yeem ntsuas qhov khoom ntawm ib qho chaw tshwj xeeb.

Koj yuav daws tau qhov ua haujlwm li cas?

Rau muaj nuj nqi , ob lub ntsiab lus txhais tau hais tias tib yam nkaus, tab sis "f (x)" muab rau koj yooj yim dua thiab cov ntaub ntawv ntau ntxiv. Koj tau hais tias "y = 2x + 3; daws rau y thaum x = –1” Tam sim no koj hais "f (x) = 2x + 3; nrhiav f (–1)" (hais tias "f-ntawm-x sib npaug 2x ntxiv rau peb; nrhiav f-of-negative-one").

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