Cov Lus Nug Piv Txwv thiab Kev Sib Tham Txog Kev Ua Haujlwm
Kev sib xyaw ua ke ntawm cov haujlwm yog lub tswv yim hauv kev lej uas ob qho kev ua haujlwm raug muab tso ua ke rau hauv ib qho. Yog tias \( f \) thiab \( g \) yog ob qho kev ua haujlwm, ces kev sib xyaw ua ke ntawm \( f \) thiab \( g \) yog ib qho kev ua haujlwm tshiab txhais ua \( (f \circ g)(x) \) uas txhais tau tias \( f(g(x)) \). Hauv tsab xov xwm no, peb yuav tham txog ntau qhov piv txwv ntawm cov teeb meem thiab yuav daws lawv li cas ntsig txog kev sib xyaw ua ke ntawm cov haujlwm.
1. Kev Nkag Siab Yooj Yim Txog Kev Ua Haujlwm ntawm Cov Khoom Siv
Ua ntej peb nkag mus rau cov lus nug piv txwv, cia peb nkag siab luv luv txog kev ua haujlwm ntawm cov khoom.
Xav tias muaj ob lub luag haujlwm \(f\) thiab \(g\):
– Muaj nuj nqi \( f \) : \( x \mapsto f(x) \)
– Muaj nuj nqi \( g \) : \( x \mapsto g(x) \)
Qhov sib xyaw ua ke ntawm \( f \) thiab \( g \), sau ua \( f \circ g \), yog ib qho kev ua haujlwm uas ua tiav:
\[ (f\circ g)(x) = f(g(x)) \]
Ntawm no, \(g(x) \) yog qhov nkag mus rau hauv lub luag haujlwm \(f \).
2. Piv txwv lus nug 1
Lo lus nug:
Muab lub luag haujlwm \( f(x) = 2x + 3 \) thiab lub luag haujlwm \( g(x) = x – 5 \). Txheeb xyuas \( (f \circ g)(x) \) thiab \( (g \circ f)(x) \).
Kev Sib Tham:
Cia peb xam thawj qhov sib xyaw ua ke \( (f \circ g)(x) \):
\[ (f\circ g)(x) = f(g(x)) \]
Kauj ruam thawj zaug, peb ntxig \( g(x) \) rau hauv \( f(x) \):
\[ g(x) = x – 5 \]
\[ f(g(x)) = f(x – 5) \]
Kauj ruam thib ob, peb sau \( x – 5 \) rau hauv lub luag haujlwm \( f \):
\[ f(x – 5) = 2(x – 5) + 3 \]
\[ = 2x – 10 + 3 \]
\[ = 2x – 7 \]
Yog li, (f \circ g)(x) = 2x - 7\).
Tam sim no cia peb xam qhov sib xyaw ua ke thib ob \( (g \circ f)(x) \):
\[ (g \circ f)(x) = g(f(x)) \]
Kauj ruam thawj zaug, peb ntxig \( f(x) \) rau hauv \( g(x) \):
\[ f(x) = 2x + 3 \]
\[ g(f(x)) = g(2x + 3) \]
Kauj ruam thib ob, peb ntxig \(2x + 3\) rau hauv lub luag haujlwm \(g\):
\[ g(2x + 3) = (2x + 3) – 5 \]
\[ = 2x + 3 – 5 \]
\[ = 2x – 2 \]
Yog li, (g\circ f)(x) = 2x - 2\).
3. Piv txwv lus nug 2: Kev sib xyaw ua ke ntawm cov haujlwm nrog Quadratic Functions
Lo lus nug:
Muab qhov kev ua haujlwm \( f(x) = x^2 + 1 \) thiab qhov kev ua haujlwm \( g(x) = 3x - 4 \). Txheeb xyuas \( (f \circ g)(x) \) thiab \( (g \circ f)(x) \).
Kev Sib Tham:
Cia peb xam thawj qhov sib xyaw ua ke \( (f \circ g)(x) \):
\[ (f\circ g)(x) = f(g(x)) \]
Kauj ruam thawj zaug, peb ntxig \( g(x) \) rau hauv \( f(x) \):
\[ g(x) = 3x – 4 \]
\[ f(g(x)) = f(3x – 4) \]
Kauj ruam thib ob, peb sau \( 3x – 4 \) rau hauv lub luag haujlwm \( f \):
\[ f(3x – 4) = (3x – 4)^2 + 1 \]
\[ = (3x – 4)(3x – 4) + 1 \]
\[ = 9x^2 – 12x ∑2 + 16 + 1 \]
\[ = 9x^2 – 24x + 16 + 1 \]
\[ = 9x^2 – 24x + 17 \]
(f \circ g)(x) = 9x^2 – 24x + 17 \).
Tam sim no cia peb xam qhov sib xyaw ua ke thib ob \( (g \circ f)(x) \):
\[ (g \circ f)(x) = g(f(x)) \]
Kauj ruam thawj zaug, peb ntxig \( f(x) \) rau hauv \( g(x) \):
\[ f(x) = x^2 + 1 \]
\[ g(f(x)) = g(x^2 + 1) \]
Kauj ruam thib ob, peb sau \( x^2 + 1 \) rau hauv lub luag haujlwm \( g \):
\[ g(x^2 + 1) = 3(x^2 + 1) – 4 \]
\[ = 3x^2 + 3 – 4 \]
\[ = 3x^2 – 1 \]
Yog li, (g \circ f)(x) = 3x^2 - 1\).
4. Piv txwv lus nug 3: Kev sib xyaw ua ke ntawm Trigonometric Functions
Lo lus nug:
Muab qhov kev ua haujlwm \( f(x) = \sin x \) thiab qhov kev ua haujlwm \( g(x) = x^2 \). Txheeb xyuas \( (f \circ g)(x) \) thiab \( (g \circ f)(x) \).
Kev Sib Tham:
Cia peb xam thawj qhov sib xyaw ua ke \( (f \circ g)(x) \):
\[ (f\circ g)(x) = f(g(x)) \]
Kauj ruam thawj zaug, peb ntxig \( g(x) \) rau hauv \( f(x) \):
\[ g(x) = x^2 \]
\[ f(x)) = f(x^2) \]
Kauj ruam thib ob, peb sau \( x^2 \) rau hauv lub luag haujlwm \( f \):
\[ f(x^2) = \sin(x^2) \]
Yog li, \( (f \circ g)(x) = \sin (x^2) \).
Tam sim no cia peb xam qhov sib xyaw ua ke thib ob \( (g \circ f)(x) \):
\[ (g \circ f)(x) = g(f(x)) \]
Kauj ruam thawj zaug, peb ntxig \( f(x) \) rau hauv \( g(x) \):
\[ f(x) = \sin x \]
\[ g(f(x)) = g(\sin x) \]
Kauj ruam thib ob, peb ntxig \( \sin x \) rau hauv qhov kev ua haujlwm \( g \):
\[ g(\sin x) = (\sin x)^2 \]
\[ = \sin^2 x \]
Yog li, \( (g \circ f)(x) = \sin^2 x \).
Xaus
Kev sib sau ua ke ntawm cov haujlwm yog ib txoj hauv kev los muab ob lub luag haujlwm sib xyaw ua ke rau hauv ib qho haujlwm. Los ntawm cov piv txwv saum toj no, peb tau kawm tias cov txheej txheem ntawm kev sib sau ua ke ntawm cov haujlwm suav nrog kev hloov ib qho haujlwm rau lwm qhov. Qhov tshwm sim kawg ntawm kev sib sau ua ke ntawm cov haujlwm nyob ntawm qhov kev txiav txim uas cov haujlwm raug siv ua ntej.
Nws yog ib qho tseem ceeb kom nkag siab tias \( (f \circ g)(x) \) tsis zoo ib yam li \( (g \circ f)(x) \), thiab qhov sib txawv no tuaj yeem tseem ceeb heev rau ntau yam kev siv lej thiab kev tshawb fawb. Yog li ntawd, kev nkag siab txog cov hauv paus thiab yuav ua li cas xam cov qauv ntawm cov haujlwm yog qhov tseem ceeb heev rau txhua tus neeg uas kawm lej ntawm qib nruab nrab lossis qib siab.
Vam tias cov kev sib tham thiab cov lus nug piv txwv saum toj no yuav muaj txiaj ntsig thiab pab cov nyeem ntawv nkag siab txog cov qauv ntawm cov haujlwm.
