MinMaxValues - 3-masala yechimi | MathModel.uz

4.3-bo'lim : Minimal va maksimal qiymatlar

3. $g\left( x \right) = {x^2} - 4x$ grafigini chizing va quyidagi oraliqlarning har birida funksiyaning barcha nisbiy ekstremal va mutlaq ekstremallarini aniqlang.

$\left( { - \infty ,\infty } \right)$
$\left[ { - 1,4} \right]$
$\left[ {1,3} \right]$
$\left[ {3,5} \right]$
$\left( { - 1,5} \right]$

a $\left( { - \infty ,\infty } \right)$ Yechim

Mana intervaldagi funksiyaning grafigi.

$The graph of $g\left( x \right) = {x^2} - 4x$ on the interval $\left( { - \infty ,\infty } \right)$. It is an upward opening parabola with vertex at $\left( {2, - 4} \right)$ that crosses the x-axis at the origin and at $\left( {4,0} \right)$. Arrows on both ends of the graph show that it continues to increase without bound in both directions.$

Agar siz parabolalarni qanday chizishni eslamasangiz, bo'limni ko'rib chiqing parabolalar grafigini tuzish Algebra eslatmalarida.

Shunday qilib, $\left( { - \infty ,\infty } \right)$ oralig'ida biz borligini aniq ko'rishimiz mumkin mutlaq maksimallar yo'q (grafik grafikning chap va o'ng tomonida chegaralarsiz ortadi.). Shuningdek bor nisbiy maksimallar yo'q (nuqta atrofidagi hududda grafik maksimal bo'lgan "zarbalar" yo'q). $\left( {2, - 4} \right)$ nuqtasi ikkalasi ham a nisbiy minimal va an mutlaq minimal.

b $\left[ { - 1,4} \right]$ Yechim

Mana bu oraliqdagi funksiyaning grafigi.

$The graph of $g\left( x \right) = {x^2} - 4x$ on the interval $\left[ { - 1,4} \right]$. It is the portion of the upward opening parabola between the two endpoints, with vertex at $\left( {2, - 4} \right)$. The graph starts at the closed endpoint $\left( { - 1,5} \right)$, falls to the vertex and then rises to the closed endpoint $\left( {4,0} \right)$.$

$\left( {2, - 4} \right)$ nuqtasi hali ham a nisbiy minimal va an mutlaq minimal. Hali ham bor nisbiy maksimallar yo'q. Biroq, biz hozir yopiq intervalda ishlayotganimiz uchun (i.e. biz chekli so'nggi nuqtalar bilan intervalda ishlayapmiz va biz oxirgi nuqtalarni o'z ichiga olamiz) bizda borligini ko'rishimiz mumkin mutlaq maksimal $\left( { - 1,5} \right)$ nuqtasida.

c $\left[ {1,3} \right]$ Yechim

Mana bu oraliqdagi funksiyaning grafigi.

$The graph of $g\left( x \right) = {x^2} - 4x$ on the interval $\left[ {1,3} \right]$. It is a short piece of the upward opening parabola running from the closed endpoint $\left( {1, - 3} \right)$, down to the vertex at $\left( {2, - 4} \right)$ and back up to the closed endpoint $\left( {3, - 3} \right)$. The whole piece lies below the x-axis.$

$\left( {2, - 4} \right)$ nuqtasi hali ham a nisbiy minimal va an mutlaq minimal. Hali ham bor nisbiy maksimallar yo'q Ushbu oraliqdagi funktsiya. Biroq, biz hozir yopiq intervalda ishlayotganimiz uchun (i.e. biz chekli so'nggi nuqtalar bilan intervalda ishlayapmiz va biz oxirgi nuqtalarni o'z ichiga olamiz) bizda borligini ko'rishimiz mumkin mutlaq maksimal Bu $\left( {1, - 3} \right)$ va $\left( {3, - 3} \right)$ nuqtalarida sodir bo'ladi.

Eslatib o'tamiz, funktsiyaning faqat bitta mutlaq maksimal qiymati (yoki shunday bo'lsa, minimal qiymat) bo'lishi mumkin bo'lsa-da, u bir nechta nuqtada paydo bo'lishi mumkin.

d $\left[ {3,5} \right]$ Yechim

Mana bu oraliqdagi funksiyaning grafigi.

$The graph of $g\left( x \right) = {x^2} - 4x$ on the interval $\left[ {3,5} \right]$. It is the piece of the upward opening parabola to the right of the vertex, so the graph increases the whole way from the closed endpoint $\left( {3, - 3} \right)$, through $\left( {4,0} \right)$, up to the closed endpoint $\left( {5,5} \right)$.$

Ushbu oraliqda bizda intervalning ichki qismida hech qanday "burilishlar" yo'q va shuning uchun bu oraliq uchun mavjud. nisbiy ekstremal yo'q Ushbu oraliqdagi funktsiya. Biroq, biz yopiq intervalda ishlamoqdamiz va shuning uchun biz borligini aniq ko'rishimiz mumkin mutlaq maksimal $\left( {5,5} \right)$ nuqtasida va an mutlaq minimal $\left( {3, - 3} \right)$ nuqtasida.

e $\left( { - 1,5} \right]$ Yechim

Mana bu oraliqdagi funksiyaning grafigi.

$The graph of $g\left( x \right) = {x^2} - 4x$ on the interval $\left( { - 1,5} \right]$. It is the portion of the upward opening parabola between the two endpoints, with vertex at $\left( {2, - 4} \right)$. There is an open dot at the left endpoint $\left( { - 1,5} \right)$, showing that it is not included, and a closed dot at the right endpoint $\left( {5,5} \right)$, showing that it is included.$

$\left( {2, - 4} \right)$ nuqtasi ikkalasi ham a nisbiy minimal va an mutlaq minimal. bor nisbiy maksimallar yo'q Ushbu oraliqdagi funktsiya.

Mutlaq maksimal uchun biz biroz ehtiyot bo'lishimiz kerak. Bunday holda, biz oraliqning o'ng oxirgi nuqtasini qo'shamiz, lekin chap tomonni emas. Shuning uchun, mavjud mutlaq maksimal $\left( {5,5} \right)$ nuqtasida. Biroq, chap nuqtada mutlaq maksimal yo'q, chunki bu nuqta intervalga kiritilmagan.

Chunki biz chap oxirgi nuqtani intervalga kiritmayapmiz va shuning uchun $x$ $x = - 1$ ga yetib bormasdan $x = - 1$ ga tobora yaqinlashib boradi. Bu shuni anglatadiki, grafik $y = 5$ ga tobora yaqinlashsa-da, u hech qachon $y = 5$ ga etib bormaydi va shuning uchun chap oxirgi nuqtada mutlaq maksimal bo'lmaydi.