Ngā tauira pātai e matapaki ana i ngā Wāhanga Tino Pai o te Uara Whakahoki Iti me te Uara Whakahoki Mōrahi

Tauira Pātai me te Kōrero mō ngā Pūwāhi Tino Rahi: Te Uara Whakahoki Iti rawa me te Uara Whakahoki Mōrahi

Ko te whakatau i ngā pūwāhi tino nui, arā, ngā pūwāhi e tae atu ai tētahi mahi ki tōna uara iti rawa, mōrahi rānei, he ariā matua tēnei i roto i te tātaitai me te tātari pāngarau. I roto i tēnei tuhinga, ka tūhuratia e mātou me pēhea te kimi me te tātari i ngā pūwāhi tino nui mā roto i ētahi tauira rapanga e pā ana ki ngā uara whakahoki iti rawa me te mōrahi.

Ngā Whakamāramatanga Kaupapa me ngā Ariā

I mua i te matapaki i ngā tauira rapanga, me mārama tātou ki ētahi ariā me ngā ariā matua:

1. Pūwāhi Aromatawai: Ko te uara o \( x \) ina kore, kāore rānei i te tīariari te pānga tuatahi \( f'(x) \) o te mahi \( f(x) \).
2. Uara Whakahoki Mōrahi: Ko te uara o \( f(x) \) he nui ake i te uara o \( f(x) \) huri noa i taua pūwāhi.
3. Uara Whakahoki Mōkito: Ko te uara o \( f(x) \) he iti iho i te uara o \( f(x) \) huri noa i taua pūwāhi.
4. Te Ariā a Fermat: Mena he uara tino ā-rohe tō \( f \) kei \( c \) ā, kei reira te tauwehenga \( f'(c) \), ko \( f'(c) = 0 \).

Tauira Pātai 1: Mahi Tapawhā

Tuatahi, ka tīmata tātou me tētahi mahi tapawhā māmā:

\[ f(x) = 2x^2 – 4x + 1 \]

Ngā Hipanga:

1. Kimihia te taupū tuatahi o \( f'(x) \):
\[
f'(x) = \frac{d}{dx}(2x^2 – 4x + 1) = 4x – 4
\]

2. Kimihia ngā pūwāhi matua mā te whakaoti rapanga \( f'(x) = 0 \):
\[
4x – 4 = 0 \te tikanga x = 1
\]

3. Whakatauhia te uara o te mahi i te pūwāhi matua:
\[
f(1) = 2(1)^2 – 4(1) + 1 = -1
\]

4. Whakamahia te tuarua o ngā tauwhitinga hei whakatau i te āhua o te pūwāhi:
\[
f”(x) = \frac{d}{dx}(4x – 4) = 4
\]
Nā te mea ko te \( f”(1) > 0 \), ko te pūwāhi \( x = 1 \) he pūwāhi iti rawa ā-rohe.

Tauira Pātai 2: Ngā Mahi Pūrau

Nā, me whakamātau tātou ki tētahi mahi pūrau uaua ake:

\[ g(x) = x^3 – 3x^2 + 2 \]

Ngā Hipanga:

1. Whakatauhia te pānga tuatahi \( g'(x) \):
\[
g'(x) = \frac{d}{dx}(x^3 – 3x^2 + 2) = 3x^2 – 6x
\]

2. Kimihia ngā pūwāhi matua mā te whakaoti rapanga \( g'(x) = 0 \):
\[
3x^2 – 6x = 0 \te tikanga 3x(x – 2) = 0 \te tikanga x = 0 \text{ or } x = 2
\]

3. Whakatauhia te uara o te mahi i te pūwāhi matua:
\[
g(0) = 0^3 – 3(0)^2 + 2 = 2
\]
\[
g(2) = 2^3 – 3(2)^2 + 2 = -2
\]

4. Whakamahia te tuarua o ngā tauwhitinga hei whakatau i te āhua o te pūwāhi:
\[
g”(x) = \frac{d}{dx}(3x^2 – 6x) = 6x – 6
\]
\[
g”(0) = 6(0) – 6 = -6 \quad (\text{uara mōrahi ā-rohe})
\]
\[
g”(2) = 6(2) – 6 = 6 \quad (\text{uara iti rawa o te rohe})
\]

Nō reira, ko te mōrahi ā-rohe o \( g(x) \) kei \( x = 0 \) me te mōrahi ā-rohe kei \( x = 2 \).

Tauira Pātai 3: Ngā Mahi Whakawhitiwhiti

Me titiro tātou ki tētahi mahi e whakamahi ana i te taupūtanga:

\[ h(x) = xe^{-x} \]

Ngā Hipanga:

1. Whakatauhia te pānga tuatahi \( h'(x) \):
\[
h'(x) = \frac{d}{dx}(xe^{-x}) = e^{-x} – xe^{-x} = (1 – x)e^{-x}
\]

2. Kimihia te pūwāhi matua mā te whakaoti rapanga \( h'(x) = 0 \):
\[
(1 – x)e^{-x} = 0 \implies 1 – x = 0 \implies x = 1
\]

3. Whakatauhia te uara o te mahi i te pūwāhi matua:
\[
h(1) = 1e^{-1} = \frac{1}{e}
\]

4. Whakamahia te tuarua o ngā tauwhitinga hei whakatau i te āhua o te pūwāhi:
\[
h”(x) = \frac{d}{dx}((1 – x)e^{-x}) = -e^{-x} – (1 – x)e^{-x} = (x – 2)e^{-x}
\]
\[
h”(1) = (1 – 2)e^{-1} = -\frac{1}{e}
\]
Nā te mea ko \( h”(1) < 0 \), ko te pūwāhi \( x = 1 \) he mōrahi ā-rohe. Tauira Raru 4: Ngā Mahi Whaitake Hei whakamutunga, ka arotakehia e mātou te mahi whaitake: \[ k(x) = \frac{x^2 + 2x}{x - 1} \]

Ngā Hipanga: 1. Kimihia te taupū tuatahi mā te whakamahi i te ture hauwhā: \[ k'(x) = \frac{(2x+2)(x-1) - (x^2+2x)}{(x-1)^2} = \frac{2x^2 - 2x + 2x - 2 - x^2 - 2x}{(x-1)^2} = \frac{x^2 - 2}{(x-1)^2} \] 2. Kimihia ngā pūwāhi arohaehae mā te whakaoti \( k'(x) = 0 \): \[ \frac{x^2 - 2}{(x-1)^2} = 0 \implies x^2 - 2 = 0 \implies x = \pm \sqrt{2} \] 3. Kimihia te uara o te mahi i ngā pūwāhi arohaehae: \[ k(\sqrt{2}) = \frac{(\sqrt{2})^2 + 2\sqrt{2}}{\sqrt{2} - 1} = \frac{2 + 2\sqrt{2}}}{\sqrt{2} - 1} \times \frac{\sqrt{2} + 1}{\sqrt{2} + 1} = \frac{(2 + 2\sqrt{2})(\sqrt{2} + 1)}{1} = 4 + 4\sqrt{2} \] \[ k(-\sqrt{2}) = \frac{(-\sqrt{2})^2 + 2(-\sqrt{2})}{-\sqrt{2} - 1} = \frac{2 - 2\sqrt{2}}{-\sqrt{2} - 1} \times \frac{-\sqrt{2} + 1}{-\sqrt{2} + 1} = \frac{(2 - 2\sqrt{2})(-\sqrt{2} + 1)}{1} = -4 + 4\sqrt{2} \] 4. Whakamahia te tātaitanga tuarua hei tirotiro i te āhua o te pūwāhi: \[ k''(x) = \frac{d}{dx}\left( \frac{x^2 - 2}{(x-1)^2} \right) \] Ka taea te mahi tātaitanga anō mā te wehewehe anō i te \( k'(x) \) e whakaatu ana mēnā ko te \( x = \sqrt{2} \) me te \( x = -\sqrt{2} \) he mōrahi ā-rohe, he mōrahi iti rānei. Whakamutunga I roto i tēnei tuhinga, kua matapakihia e mātou ētahi tauira e whakaatu ana me pēhea te kimi i te tino, arā, ko ngā uara whakamuri iti rawa me te mōrahi, o ngā momo mahi rerekē. Ko ngā tikanga e whakamahia ana ko te kimi i te tātaitanga tuatahi hei kimi i ngā pūwāhi tino nui, te whakamahi i te tātaitanga tuarua hei whakatau i te āhua o ngā pūwāhi, me te aromatawai i te mahi i aua pūwāhi. He tūāpapa pakari tēnei mō te ruku hohonu ake ki te tātari mahi i roto i te tātaitai.