Tambayoyi da Tattaunawa kan Ƙara Ayyuka, Rage Ayyuka, da Ayyukan Tsaftacewa
Ayyukan lissafi batu ne mai zurfi kuma yana cike da halaye daban-daban, ɗaya daga cikinsu shine yadda za a iya nazarin su dangane da ƙaruwa, raguwa, ko yanayin da ba ya tsayawa. Sanin ko aiki yana ƙaruwa, raguwa, ko kuma yana tsayawa a wani lokaci yana da mahimmanci a aikace-aikacen lissafi daban-daban, gami da tattalin arziki, kimiyyar lissafi, da injiniyanci. Wannan labarin zai rufe misalai da tattaunawarsu da ta shafi ƙaruwa, raguwa, da ayyukan da ba sa tsayawa.
Menene ayyuka masu ƙaruwa, ayyuka masu raguwa, da ayyuka marasa tsayawa?
1. Ƙaruwar Aiki: Ana cewa aikin \(f(x) \) yana ƙaruwa akan tazara \(I \) idan ga kowane \(x_1 \) da \(x_2 \) a cikin \(I \) tare da \(x_1 <x_2 \), muna da \(f(x_1) \leq f(x_2) \). 2. Rage Aikin: Akasin haka, ana cewa aikin \(f(x) \) yana raguwa akan tazara \(I \) idan ga kowane \(x_1 \) da \(x_2 \) a cikin \(I \) tare da \(x_1 <x_2 \), muna da \(f(x_1) \geq f(x_2) \). 3. Aikin Tsaftacewa: Ana cewa aiki \( f(x) \) yana tsayawa akan tazara \( I \) idan ga kowane \( x \) a cikin \( I \), aikin yana da ƙima iri ɗaya, wato \( f(x) = c \) ga kowane \( x \) a cikin \( I \), inda c yake dindindin. Misali Tambaya ta 1: Ƙayyade Tazara na Aikin Ƙaruwa Ganin cewa aikin \( f(x) = 2x^3 - 3x^2 - 12x + 5 \). Ƙayyade tazara inda aikin ke ƙaruwa! Tattaunawa: Don ƙayyade tazara inda aikin ke ƙaruwa, muna buƙatar nemo farkon abin da aikin ya samo asali sannan mu bincika alamar abin da ya samo asali. 1. Mataki na 1: Nemo asalin farko: \[f'(x) = d/dx (2x^3 - 3x^2 - 12x + 5) \] \[f'(x) = 6x^2 - 6x - 12 \] 2. Mataki na 2: Ƙayyade mahimman bayanai: Muhimman bayanai sune wuraren da asalin farko sifili ne ko kuma ba a bayyana shi ba. \[ 6x^2 - 6x - 12 = 0 \] Raba dukkan daidaiton da 6: \[ x^2 - x - 2 = 0 \] Mun ƙididdige wannan daidaiton kwata: \[ (x-2)(x+1) = 0 \] Don haka, mahimman bayanai sune \( x = 2 \) da \( x = -1 \). 3. Mataki na 3: Kayyade alamar farkon abin da aka samo a kan tazara da aka samar ta hanyar mahimman bayanai: Za mu ƙirƙiri teburin alama don \( f'(x) \) akan tazara \( (-\infty, -1) \), \( (-1, 2) \), da \( (2, \infty) \). - Ga \( x \in (-\infty, -1) \): Ɗauki \( x = -2 \) \[ f'(-2) = 6(-2)^2 - 6(-2) - 12 = 24 + 12 - 12 = 24 \] Tunda \( f'(-2) > 0 \), to \( f(x) \) yana ƙaruwa akan tazara \( (-\infty, -1) \).
– Domin \( x \in (-1, 2) \): Ɗauki \( x = 0 \)
\[ f'(0) = 6(0)^2 – 6(0) – 12 = -12 \]
Tunda \( f'(0) < 0 \), to \( f(x) \) yana raguwa akan tazara \(-1, 2) \). - Don \( x \in (2, \infty) \): Ɗauki \( x = 3 \) \[ f'(3) = 6(3)^2 - 6(3) - 12 = 54 - 18 - 12 = 24 \] Tunda \( f'(3) > 0 \), to \( f(x) \) yana ƙaruwa akan tazara \( (2, \infty) \).
Don haka, aikin \( f(x) \) yana ƙaruwa akan tazara \((-\infty, -1) \cup (2, \infty) \).
Misali Matsala ta 2: Tantance Tazarar Aikin Ragewa
Idan aka yi la'akari da aikin \( g(x) = 4x^4 – 8x^3 + 2 \). Ka ƙayyade tazara inda aikin ke raguwa!
Tattaunawa:
1. Mataki na 1: Nemo asalin farko:
\[ g'(x) = d/dx (4x^4 – 8x^3 + 2) \]
\[ g'(x) = 16x^3 – 24x^2 \]
2. Mataki na 2: Kayyade muhimmin batu:
\[ 16x^3 – 24x^2 = 0 \]
\[ 8x^2(2x – 3) = 0 \]
Don haka muhimman abubuwan sune \( x = 0 \) da \( x = \frac{3}{2} \).
3. Mataki na 3: Kayyade alamar farkon abin da aka samo a lokacin:
– Domin \( x \in (-\infty, 0) \): Ɗauki \( x = -1 \)
\[ g'(-1) = 16(-1)^3 – 24(-1)^2 = -16 – 24 = -40 \]
Tunda \( g'(-1) < 0 \), to \( g(x) \) yana raguwa akan tazara \( (-\infty, 0) \). - Don \( x \in (0, \frac{3}{2}) \): Ɗauki \( x = 1 \) \[ g'(1) = 16(1)^3 - 24(1)^2 = 16 - 24 = -8 \] Tunda \( g'(1) < 0 \), to \( g(x) \) yana raguwa akan tazara \( (0, \frac{3}{2}) \). - Domin \( x \in (\frac{3}{2}, \infty) \): Ɗauki \( x = 2 \) \[ g'(2) = 16(2)^3 - 24(2)^2 = 128 - 96 = 32 \] Tunda \( g'(2) > 0 \), to \( g(x) \) yana ƙaruwa akan tazara \((\frac{3}{2}, \infty) \).
Don haka, aikin \( g(x) \) yana raguwa akan tazara \( (-\infty, 0) \cup (0, \frac{3}{2}) \).
Misali Tambaya ta 3: Ƙayyade Tazarar Aiki a Hutu
Idan aka yi la'akari da aikin \( h(x) = 7 \), a ƙayyade tazara inda aikin yake a tsaye!
Tattaunawa:
Aiki mai dorewa kamar \( h(x) = 7 \) yana da asalin farko na sifili ga duka \( x \):
\[h'(x) = 0 \]
Tunda farkon wanda aka samo asali koyaushe sifili ne, aikin yana tsayawa akan dukkan yankin, don haka zamu iya cewa aikin \( h(x) = 7 \) yana tsayawa akan duk lambobi na gaske, wanda a cikin bayanin tazara shine \( (-\infty, \infty) \).
Kammalawa
Fahimtar tazara tsakanin ƙaruwa, raguwa, da kuma yanayin aiki a tsaye wani muhimmin ɓangare ne na nazarin aiki. Ta hanyar misalan da ke sama, mun rufe muhimman ra'ayoyi da matakan da ake buƙata don nemo waɗannan tazara. Wannan ilimin yana da matuƙar amfani a aikace-aikace daban-daban na lissafi da ka'idoji.
