Ngā Tauira Pātai me te Kōrero mō te Ariā o ngā Pānga Taurite
Ko te pānga o tētahi mahi he ariā taketake i roto i te tātaitai e whānuitia ana te whakamahinga i roto i ngā momo marautanga, pērā i te ahupūngao, te ōhanga, me te hangarau. Ka kapi tēnei tuhinga i ētahi tauira rapanga, ā, ka matapakihia hoki te ariā o te pānga o tētahi mahi hei whakarato i tētahi māramatanga hōhonu ake mō tēnei kaupapa.
Te Whakamāramatanga Taketake o ngā Hua Whakaputa
I mua i te urunga atu ki ngā tauira pātai, he mea pai kia arotakehia te whakamāramatanga me ngā kaupapa matua o ngā tātaitanga. Ko te tātaitanga o tētahi mahi \( f(x) \) i te pūwāhi \( x = a \) ko:
\[ f'(a) = \lim_{{h \to 0}} \frac{f(a+h) – f(a)}{h} \]
Ko te mahi \( f'(x) \) e kiia ana ko te mahi taupū o \( f(x) \).
Tauira Pātai 1: Ngā Tauwehenga Pūrau Taketake
Pātai:
Kimihia te pānga tuatahi o te mahi \( f(x) = 3x^3 – 5x^2 + 2x – 7 \).
Kōrero:
Whakamahia te ture taketake mō ngā tātaitanga \( \frac{d}{dx} x^n = nx^{n-1} \).
1. Mō \( 3x^3 \):
\[ \frac{d}{dx}(3x^3) = 3 \cdot 3x^{3-1} = 9x^2 \]
2. Mō \( -5x^2 \):
\[ \frac{d}{dx}(-5x^2) = -5 \cdot 2x^{2-1} = -10x \]
3. Mō \( 2x \):
\[ \frac{d}{dx}(2x) = 2 \]
4. Mō \( -7 \):
\[ \frac{d}{dx}(-7) = 0 \]
Pēnei:
\[ f'(x) = 9x^2 – 10x + 2 \]
Tauira Pātai 2: Ngā Pānga o ngā Mahi Pāngatoru
Pātai:
Kimihia te pānga tuatahi o te mahi \( g(x) = \sin(x) \cdot \cos(x) \).
Kōrero:
Whakamahia te ture hua \( \frac{d}{dx} [u(x) \cdot v(x)] = u'(x)v(x) + u(x)v'(x) \) me \( u(x) = \sin(x) \) me \( v(x) = \cos(x) \).
1. Ko te taupū o \( \sin(x) \) ko \( \cos(x) \), nō reira \( u'(x) = \cos(x) \).
2. Ko te pānga o \( \cos(x) \) ko \( -\sin(x) \), nō reira \( v'(x) = -\sin(x) \).
Whakakapinga \( u'(x) \) me \( v'(x) \):
\[ g'(x) = \cos(x) \cdot \cos(x) + \sin(x) \cdot (-\sin(x)) \]
\[ g'(x) = \cos^2(x) – \sin^2(x) \]
Ko te hua whakamutunga:
\[ g'(x) = \cos^2(x) – \sin^2(x) \]
Tauira 3: Te Pūtake o te Mahi Taupū
Pātai:
Kimihia te pānga tuatahi o te mahi \( h(x) = e^{2x} \).
Kōrero:
Whakamahia te ture o te pānga taupū o te mahi taupū \( \frac{d}{dx} e^{kx} = ke^{kx} \) me \( k = 2 \).
\[ h'(x) = \frac{d}{dx} e^{2x} \]
\[ h'(x) = 2 \cdot e^{2x} \]
Ko te hua whakamutunga:
\[ h'(x) = 2e^{2x} \]
Tauira Pātai 4: Te Pūtake o te Mahi Logarithmic
Pātai:
Kimihia te pānga tuatahi o te mahi \( p(x) = \ln(3x + 1) \).
Kōrero:
Whakamahia te ture o te pānga o te mahi taupūnga \( \frac{d}{dx} \ln(u) = \frac{1}{u} \cdot u' \) me \( u(x) = 3x + 1 \).
1. Kimihia te tauwehenga ā-roto \( u(x) = 3x + 1 \):
\[ u'(x) = 3 \]
2. Whakamahia te ture taupū taupū:
\[ p'(x) = \frac{1}{3x + 1} \cdot 3 \]
Ko te hua whakamutunga:
\[ p'(x) = \frac{3}{3x + 1} \]
Tauira Pātai 5: Te Whakamahinga o ngā Pānga Taurite – Te Mōrahi me te Mōkito
Pātai:
Kimihia ngā uara mōrahi me ngā uara mōkito o te mahi \( q(x) = -2x^3 + 3x^2 + 12x – 5 \) i te wā \( x \in [-2, 2] \).
Kōrero:
1. Kimihia te taupū tuatahi o \( q(x) \):
\[ q'(x) = \frac{d}{dx}(-2x^3 + 3x^2 + 12x – 5) \]
\[ q'(x) = -6x^2 + 6x + 12 \]
2. Kimihia ngā pūwāhi tūmau mā te whakaoti i te \( q'(x) = 0 \):
\[ -6x^2 + 6x + 12 = 0 \]
\[ -6(x^2 – x – 2) = 0 \]
\[ x^2 – x – 2 = 0 \]
\[ (x-2)(x+1) = 0 \]
Ko ngā pūwāhi tūmau ko \( x = 2 \) me \( x = -1 \).
3. Aromatawaihia \( q(x) \) i ngā pūwāhi matua me ngā rohe āputa:
q(-2) = -2(-2)^3 + 3(-2)^2 + 12(-2) – 5
\[ = 16 + 12 – 24 – 5 \]
\[ = -1 \]
q(2) = -2(2)^3 + 3(2)^2 + 12(2) – 5
\[ = -16 + 12 + 24 – 5 \]
\[ = 15 \]
q(-1) = -2(-1)^3 + 3(-1)^2 + 12(-1) – 5
\[ = 2 + 3 – 12 – 5 \]
\[ = -12 \]
4. Aromatawai i ngā hua:
– Ka puta te uara mōrahi i te \( x = 2 \) me te \( q(2) = 15 \).
– Ka puta te uara iti rawa i te \( x = -1 \) me te \( q(-1) = -12 \).
Te Katinga
He mea nui te māramatanga hōhonu ki te ariā o te tauwehenga o tētahi mahi i roto i ngā momo mara pūtaiao. Ko te tumanako, mā ngā tauira rapanga me ngā kōrero i runga ake nei ka āwhina i a koe ki te whakahōhonu ake i tō māramatanga ki te ariā. I roto i te mahi, he maha ngā wā me whakakotahi tātou i ngā ture me ngā ariā hei whakaoti rapanga uaua ake. Kia pai tō ako!
