Uyini umkhawulo womsebenzi

Kule ncwadi, sizocabangela omunye wemiqondo eyinhloko yokuhlaziywa kwezibalo - umkhawulo womsebenzi: incazelo yawo, kanye nezixazululo ezihlukahlukene ezinezibonelo ezisebenzayo.

Inquma umkhawulo womsebenzi

Umkhawulo womsebenzi – inani inani lalo msebenzi elithambekele kulo uma impikiswano yawo ifinyelela endaweni yokukhawulela.

Khawulela irekhodi:

• umkhawulo uboniswa isithonjana lim;
• ngezansi kwengezwe ukuthi i-agumenti (i-variable) yomsebenzi ithambekele ekubeni ngakanani. Ngokuvamile lokhu x, kodwa hhayi ngempela, isibonelo:x→1″;
• bese umsebenzi ngokwawo wengezwa kwesokudla, isibonelo:

  

Ngakho-ke, irekhodi lokugcina lomkhawulo libukeka kanje (kithi):

Ifundeka kanje "umkhawulo womsebenzi njengoba u-x ethambekele ebunyeni".

x→ 1 - lokhu kusho ukuthi u-"x" uhlala ethatha amanani asondela ngokungapheli ebunyeni, kodwa angeke aqondane nawo (ngeke afinyelelwe).

Imikhawulo yesinqumo

Ngenombolo enikeziwe

Masixazulule umkhawulo ongenhla. Ukuze wenze lokhu, mane ufake iyunithi esikhundleni somsebenzi (ngoba x→1):

Ngakho-ke, ukuxazulula umkhawulo, siqala ngokuzama ukufaka inombolo enikeziwe esikhundleni somsebenzi ongezansi kwayo (uma u-x ejwayele inombolo ethile).

Ngokungapheli

Kulokhu, ukungqubuzana komsebenzi kukhula ngokungapheli, okungukuthi, "X" ijwayele ukungapheli (∞). Ngokwesibonelo:

If x→∞, bese umsebenzi onikeziwe uthambekele ekususeni okungapheli (-∞), ngoba:

• 3 - 1 = 2
• 3 – 10 = -7
• 3 – 100 = -97
• 3 – 1000 – 997 njll.

Esinye isibonelo esiyinkimbinkimbi

Ukuze uxazulule lo mkhawulo, futhi, mane ukhulise amanani x futhi ubheke "ukuziphatha" komsebenzi kuleli cala.

• RџS•Rё x = 1, y = 1² + 3 · 1 – 6 = -2
• RџS•Rё x = 10, y = 10² + 3 · 10 – 6 = 124
• RџS•Rё x = 100, y = 100² + 3 · 100 – 6 = 10294

Ngakho-ke, ngoba "X"ukuthambekela kokungapheli, umsebenzi x² + 3x-6 ikhula ngokungapheli.

Ngokungaqiniseki (x ijwayele ukungapheli)

Kulokhu, sikhuluma ngemikhawulo, lapho umsebenzi uyingxenyana, inombolo kanye nedinominator okuyizipholynomials. Lapho "X" ijwayele ukungapheli.

Isibonelo: ake sibale umkhawulo ngezansi.

Isixazululo

Izinkulumo kukho kokubili inombolo kanye nedinominetha zivame ukungapheli. Kungacatshangwa ukuthi kulesi simo isisombululo sizoba kanje:

Nokho, akuwona wonke elula kangaka. Ukuze sixazulule umkhawulo sidinga ukwenza okulandelayo:

1. Thola x emandleni aphakeme kakhulu enombolo (kithi, amabili).

2. Ngokufanayo, siyachaza x emandleni aphakeme kakhulu edinomineyitha (futhi ilingana nokubili).

3. Manje sehlukanisa kokubili inombolo kanye nedinominetha ngayo x ngeziqu eziphezulu. Esimweni sethu, kuzo zombili izimo - kwesibili, kodwa uma behlukene, kufanele sithathe izinga eliphezulu kakhulu.

4. Emphumeleni womphumela, zonke izingxenyana zivame ukuba ziro, ngakho-ke impendulo ingu-1/2.

Ngokungaqiniseki (x ijwayele inombolo ethile)

Kokubili inombolo kanye nedinominayizi ngama-polynomials, nokho, "X" ijwayele inombolo ethile, hhayi kokungapheli.

Kulokhu, sivala amehlo ethu ngokwemibandela yokuthi i-denominator ingu-zero.

Isibonelo: Ake sithole umkhawulo womsebenzi ongezansi.

Isixazululo

1. Okokuqala, ake sishintshe inombolo yoku-1 emsebenzini, lapho "X". Sithola ukungaqiniseki kwefomu esilicabangelayo.

2. Okulandelayo, sihlukanisa inombolo nedinominetha zibe izici. Ukuze wenze lokhu, ungasebenzisa amafomula wokuphindaphinda afushanisiwe, uma efanelekile, noma.

Esimweni sethu, izimpande zenkulumo ekubaleni (2x² - 5x + 3 = 0) izinombolo 1 kanye no-1,5. Ngakho-ke, ingamelwa ngokuthi: 2(x-1)(x-1,5).

I-Denominator (x–1) ilula ekuqaleni.

3. Sithola umkhawulo onjalo oshintshiwe:

4. I-fraction ingancishiswa ngokuthi (x–1):

5. Kusele kuphela ukufaka inombolo engu-1 enkulumweni etholwe ngaphansi komkhawulo: