4.5: Kazi kamili ya Thamani
- Page ID
- 164636
Kwa grafu kazi absolute-thamani, kuchagua maadili ndogo ya\(x\), na compute thamani ya\(f(x)\) kutoka kazi kutokana na kujenga jozi kuamuru. Tatu zilizoamriwa jozi ni kiasi cha chini kinachohitajika ili graph kazi ya thamani kamili. Kuwa makini, kwa sababu jozi moja iliyoamriwa lazima iwakilisha vertex, hatua ambapo pande za kushoto na za kulia za kazi hukutana. Kwa usahihi grafu sura ya kazi ya thamani kabisa, vertex inapaswa kupatikana.
\(f(x) = a\vert x − h\vert+ k\)Fomu ya jumla ya Kazi ya Thamani kamili, na vertex\((h,k)\)
- \(a\)huamua upana na mwelekeo (inakabiliwa juu au chini) ya kazi.
- \(h\)ni mabadiliko ya usawa kutoka kwa asili.
- \(k\)ni mabadiliko ya wima kutoka kwa asili.
Anza kwa kutambua jozi iliyoamriwa ya vertex, na kisha kupata jozi moja iliyoamriwa upande wa kushoto wa asili, na kwa haki ya asili. Chagua kitengo cha x-thamani moja upande wa kushoto wa thamani ya x-ya asili, uhesabu\(f(x)\) na kisha uchague kitengo kimoja cha thamani ya x ya asili na uhesabu\(f(x)\). Grafu itafanana\(V\), ama inakabiliwa juu au chini, kulingana na ishara ya\(a\).
Unda meza ya ufumbuzi na grafu kazi ya thamani kamili:
\(f(x) = \vert x − 4\vert\)
Suluhisho
Kulinganisha kazi hii kwa fomu ya jumla ya kazi za thamani kamili (iliyoonyeshwa hapo juu),\(a = 1\),\(h = 4\),\(k = 0\). Vertex ni\((h, k)\) au\((4, 0)\).
Ili kupata mbili zaidi kuamuru jozi, kuchagua\(x = 3\) na\(x = 5\), kisha compute maadili ya\(f(x)\).
\(x\) | \(f(x)\) |
---|---|
\ (x\) ">3 | \ (f (x)\) ">\(f(3) = \vert 3 − 4\vert = \vert − 1\vert = 1\) |
\ (x\) ">4 | \ (f (x)\) ">\(f(4) = \vert 4 − 4\vert = \vert 0\vert = 0\) |
\ (x\) "> 5 | \ (f (x)\) ">\(f(5) = \vert 5 − 4\vert = \vert 1\vert = 1\) |
Unda meza ya ufumbuzi na grafu kazi ya thamani kamili:
\(g(x) = \vert x + 2\vert − 5\)
Suluhisho
Kulinganisha kazi hii kwa fomu ya jumla ya kazi za thamani kamili (iliyoonyeshwa hapo juu),\(a = 1\),\(h = −2\),\(k = −5\). Kipeo ni (h, k)\) au\((−2, −5)\).
Ili kupata jozi mbili zilizoamriwa, chagua\(x = −3\) na\(x = −1\), kisha uhesabu maadili ya\(g(x)\)
Jedwali la Solutions kwa\(g(x) = \vert x + 2\vert − 5\) | |
\(x\) | \(g(x)\) |
-3 | \(g(−3) = \vert − 3 + 2\vert − 5 = \vert − 1\vert − 5 = 1 − 5 = −4\) |
-2 | \(g(−2) = \vert − 2 + 2\vert − 5 = \vert 0\vert − 5 = 0 − 5 = −5\) |
-1 | \(g(−1) = \vert − 1 + 2\vert − 5 = \vert 1\vert − 5 = 1 − 5 = −4\) |
Unda meza ya ufumbuzi na grafu kazi ya thamani kamili:
\(h(x) = 3\vert x − 5\vert + 1\)
Suluhisho
Kulinganisha kazi hii kwa fomu ya jumla ya kazi za thamani kamili (iliyoonyeshwa hapo juu),\(a = 3\),\(h = 5\),\(k = 1\). Vertex ni\((h, k)\) au\((5, 1)\).
Ili kupata mbili zaidi kuamuru jozi, kuchagua\(x = 4\) na\(x = 6\), kisha compute maadili ya\(h(x)\).
Jedwali la Solutions kwa\(h(x) = 3\vert x − 5\vert + 1\) | |
\(x\) | \(h(x)\) |
4 | \(g(−3) = \vert − 3 + 2\vert − 5 = \vert − 1\vert − 5 = 1 − 5 = −4\) |
5 | \(g(−2) = \vert − 2 + 2\vert − 5 = \vert 0\vert − 5 = 0 − 5 = −5\) |
6 | \(g(−1) = \vert − 1 + 2\vert − 5 = \vert 1\vert − 5 = 1 − 5 = −4\) |
Unda meza ya ufumbuzi na grafu kazi ya thamani kamili:
\(h(x) = \dfrac{1}{2} \vert x − 2\vert + 3\)
Suluhisho
Kulinganisha kazi hii kwa fomu ya jumla ya kazi za thamani kamili (iliyoonyeshwa hapo juu),\(a = \dfrac{1}{2} \),\(h = 2\),\(k = 3\). Vertex ni\((h, k)\) au\((2, 3)\).
Ili kupata mbili zaidi kuamuru jozi, kuchagua\(x = 1\) na\(x = 3\), kisha compute maadili ya\(h(x)\).
Jedwali la Solutions kwa\(h(x) = \dfrac{1}{2} \vert x − 2\vert + 3\) | |
\(x\) | \(h(x)\) |
1 | \(h(1) = \dfrac{1}{2} \vert 1 − 2\vert + 3 = \dfrac{1}{2} \vert − 1\vert + 3 = \dfrac{1}{2} (1) + 3 = 3\dfrac{1}{2}\) |
2 | \(h(2) = \dfrac{1}{2} \vert 2 − 2\vert + 3 = \dfrac{1}{2} \vert 0\vert + 3 = 0 + 3 = 3\) |
3 | \(h(3) = \dfrac{1}{2} \vert 3 − 2\vert + 3 = \dfrac{1}{2} \vert 1\vert + 3 = \dfrac{1}{2} (1) + 3 = 3\dfrac{1}{2}\) |
Unda meza ya ufumbuzi na grafu kazi zifuatazo za thamani kamili:
- \(f(x) = \vert x + 6\vert\)
- \(g(x) = \dfrac{1}{3} \vert x − 3\vert + 5\)
- \(h(x) = 4\vert x + 2\vert + 2\)
- \(f(x) = \vert x − 1\vert − 5\)