Su'aalo iyo Doodo Tusaalooyin ah oo ku saabsan Dhibcaha Xad-dhaafka ah: Qiimaha Soo Celinta Ugu Yar iyo Qiimaha Soo Celinta Ugu Badan
Go'aaminta dhibcaha xad-dhaafka ah, dhibcaha uu shaqadu gaarto qiimaheeda ugu yar ama ugu badan, waa fikrad asaasi ah oo ku saabsan falanqaynta xisaabinta iyo xisaabta. Maqaalkan, waxaan ku sahamin doonnaa sida loo helo loona falanqeeyo dhibcaha xad-dhaafka ah iyada oo loo marayo dhowr dhibaato oo tusaale ah oo ku lug leh qiimaha ugu yar iyo kan ugu badan ee soo celinta.
Qeexitaannada Mawduuca iyo Aragtiyada
Kahor inta aan laga wada hadlin dhibaatooyinka tusaalaha ah, waxaan u baahanahay inaan fahanno qaar ka mid ah fikradaha aasaasiga ah iyo aragtiyaha:
1. Barta Muhiimka ah: Waa qiimaha \( x \) halkaas oo beddelka ugu horreeya ee \( f'(x) \) ee shaqada \( f(x) \) uu eber yahay ama uusan jirin.
2. Qiimaha Soo Celinta Ugu Badan: Waa qiimaha \( f(x) \) ee ka weyn qiimaha \( f(x) \) ee ku wareegsan bartaas.
3. Qiimaha ugu Yar ee Soo Celinta: Waa qiimaha \( f(x) \) ee ka yar qiimaha \( f(x) \) ee ku wareegsan bartaas.
4. Aragtida Fermat: Haddii \( f \) uu leeyahay qiime xad-dhaaf ah oo maxalli ah oo ku yaal \( c \) iyo ka-soo-horjeedka \( f'(c) \) uu jiro, markaas \( f'(c) = 0 \).
Su'aal Tusaale 1: Shaqooyinka Labajibbaaran
Marka hore, waxaan ku bilaabaynaa shaqo labajibbaaran oo fudud:
\[ f(x) = 2x^2 – 4x + 1 \]
Talaabooyinka:
1. Soo hel asalka ugu horreeya ee \( f'(x) \):
\[
f'(x) = \frac{d}{dx}(2x^2 – 4x + 1) = 4x – 4
\]
2. Soo hel qodobbada muhiimka ah adoo xallinaya \( f'(x) = 0 \):
\[
4x – 4 = 0 \waxay la macno tahay x = 1
\]
3. Go'aami qiimaha shaqada marka la eego barta muhiimka ah:
\[
f(1) = 2(1)^2 – 4(1) + 1 = -1
\]
4. Adeegso nooca labaad ee wax laga beddelay si aad u go'aamiso nooca dhibicda:
\[
f”(x) = \frac{d}{dx}(4x – 4) = 4
\]
Maadaama \( f”(1) > 0 \), dhibicda \( x = 1 \) ay tahay dhibicda ugu yar ee maxalliga ah.
Su'aal Tusaale 2: Shaqooyinka Polynomial
Hadda aan isku dayno shaqo polynomial ah oo aad u adag:
\[ g(x) = x^3 – 3x^2 + 2 \]
Talaabooyinka:
1. Go'aami nooca ugu horreeya ee ka soo jeeda \( g'(x) \):
\[
g'(x) = \frac{d}{dx}(x^3 – 3x^2 + 2) = 3x^2 – 6x
\]
2. Soo hel qodobbada muhiimka ah adoo xallinaya \( g'(x) = 0 \):
\[
3x^2 – 6x = 0 \waxay tilmaamaysaa 3x(x – 2) = 0 \waxay tilmaamaysaa x = 0 \qoraalka{ ama } x = 2
\]
3. Go'aami qiimaha shaqada marka la eego barta muhiimka ah:
\[
g(0) = 0^3 – 3(0)^2 + 2 = 2
\]
\[
g(2) = 2^3 – 3(2)^2 + 2 = -2
\]
4. Adeegso nooca labaad ee wax laga beddelay si aad u go'aamiso nooca dhibicda:
\[
g”(x) = \frac{d}{dx}(3x^2 – 6x) = 6x – 6
\]
\[
g”(0) = 6(0) – 6 = -6 \quad (\text{qiimaha ugu badan ee maxalliga ah})
\]
\[
g”(2) = 6(2) – 6 = 6 \quad (\text{qiimaha ugu yar ee maxalliga ah})
\]
Markaa, \( g(x) \) wuxuu leeyahay ugu badnaan maxalli ah oo ah \( x = 0 \) iyo ugu yaraan maxalli ah oo ah \( x = 2 \).
Su'aal Tusaale ah 3: Hawlaha Transcendental
Aan eegno shaqo ku lug leh erey-bixin:
\[ h(x) = xe^{-x} \]
Talaabooyinka:
1. Go'aami nooca ugu horreeya ee ka soo jeeda \( h'(x) \):
\[
h'(x) = \frac{d}{dx}(xe^{-x}) = e^{-x} – xe^{-x} = (1 – x) e^{-x}
\]
2. Soo hel qodobka muhiimka ah adoo xallinaya \( h'(x) = 0 \):
\[
(1 – x)e^{-x} = 0 \waxay tilmaamaysaa 1 – x = 0 \waxay tilmaamaysaa x = 1
\]
3. Go'aami qiimaha shaqada marka la eego barta muhiimka ah:
\[
h(1) = 1e^{-1} = \frac{1}{e}
\]
4. Adeegso nooca labaad ee wax laga beddelay si aad u go'aamiso nooca dhibicda:
\[
h”(x) = \frac{d}{dx}((1 – x)e^{-x}) = -e^{-x} – (1 – x)e^{-x} = (x – 2)e^{-x}
\]
\[
h” (1) = (1 – 2)e^{-1} = -\frac{1}{e}
\]
Maadaama \( h”(1) < 0 \), dhibicda \( x = 1 \) ay tahay ugu badnaan maxalli ah. Tusaale ahaan Dhibaatada 4: Hawlaha Caqliga leh Ugu dambeyntii, waxaan qiimeyneynaa shaqada caqliga leh: \[ k(x) = \frac{x^2 + 2x}{x - 1} \] Tallaabooyinka: 1. Soo hel beddelka ugu horreeya adoo isticmaalaya xeerka saamiga: \[ k'(x) = \frac{(2x+2)(x-1) - (x^2+2x)}{(x-1)^2} = \frac{2x^2 - 2x + 2x - 2 - x^2 - 2x}{(x-1)^2} = \frac{x^2 - 2}{(x-1)^2} \] 2. Soo hel qodobbada muhiimka ah adoo xallinaya \( k'(x) = 0 \): \[ \frac{x^2 - 2}{(x-1)^2} = 0 \tilmaamaya x^2 - 2 = 0 \tilmaamaya x = \pm \sqrt{2} \] 3. Go'aami qiimaha shaqada barta muhiimka ah: \[ k(\sqrt{2}) = \frac{(\sqrt{2})^2 + 2\sqrt{2}}{\sqrt{2} - 1} = \frac{2 + 2\sqrt{2}}{\sqrt{2} - 1} \times \frac{\sqrt{2} + 1}{\sqrt{2} + 1} = \frac{(2 + 2\sqrt{2})(\sqrt{2} + 1)}{1} = 4 + 4\sqrt{2} \] \[ k(-\sqrt{2}) = \frac{(-\sqrt{2})^2 + 2(-\sqrt{2})}{-\sqrt{2} - 1} = \frac{2 - 2\sqrt{2}}{-\sqrt{2} - 1} \times \frac{-\sqrt{2} + 1}{-\sqrt{2} + 1} = \frac{(2 - 2\sqrt{2})(-\sqrt{2} + 1)}{1} = -4 + 4\sqrt{2} \] 4. Isticmaal derivative labaad si aad u hubiso sifooyinka barta: \[k''(x) = \frac{d}{dx}\left( \frac{x^2 - 2}{(x-1)^2} \right) \] Xisaabin dheeraad ah ayaa lagu samayn karaa iyadoo la kala saarayo \( k'(x) \) Natiijooyinka waxay tilmaamayaan in \( x = \sqrt{2} \) iyo \( x = -\sqrt{2} \) ay yihiin dhibcaha ugu badan ama kuwa ugu yar ee maxalliga ah. Gunaanad Maqaalkan, waxaan ka wada hadalnay dhowr dhibaato oo tusaale ah oo muujinaya sida loo helo xadka sare, taas oo ah, ugu yaraan iyo ugu badnaan rogaal celinta qiimayaasha, noocyada kala duwan ee hawlaha. Farsamooyinka la isticmaalay waxaa ka mid ah helitaanka asalka koowaad si loo helo qodobbada muhiimka ah, isticmaalka asalka labaad si loo go'aamiyo nooca dhibcaha, iyo qiimeynta shaqada meelahaas. Tani waa aasaas adag oo lagu quusto falanqaynta shaqada ee xisaabinta.
