{"id":448,"date":"2012-10-29T23:20:00","date_gmt":"2012-10-29T14:20:00","guid":{"rendered":"http:\/\/www.roundown.net\/nyushi\/?p=448"},"modified":"2021-09-16T05:59:14","modified_gmt":"2021-09-15T20:59:14","slug":"ngr201002","status":"publish","type":"post","link":"https:\/\/www.roundown.net\/nyushi\/ngr201002\/","title":{"rendered":"\u540d\u53e4\u5c4b\u5927\u7406\u7cfb2010\uff1a\u7b2c2\u554f"},"content":{"rendered":"
\n

\u95a2\u6570 \\(f(x) = (x^2-x) e^{-x}\\) \u306b\u3064\u3044\u3066, \u6b21\u306e\u554f\u3044\u306b\u7b54\u3048\u3088. \u5fc5\u8981\u306a\u3089\u3070, \u4efb\u610f\u306e\u81ea\u7136\u6570 \\(n\\) \u306b\u5bfe\u3057\u3066\r\n\\[\r\n\\displaystyle\\lim _ {x \\rightarrow + \\infty} x^n e^{-x} = 0\r\n\\]\r\n\u304c\u6210\u308a\u7acb\u3064\u3053\u3068\u3092\u7528\u3044\u3066\u3082\u3088\u3044.<\/p>\r\n

    \r\n

  1. (1)<\/strong>\u3000\\(y = f(x)\\) \u306e\u30b0\u30e9\u30d5\u306e\u5909\u66f2\u70b9\u3092\u6c42\u3081, \u30b0\u30e9\u30d5\u306e\u6982\u5f62\u3092\u304b\u3051.<\/p><\/li>\r\n

  2. (2)<\/strong>\u3000\\(a \\gt 0\\) \u3068\u3059\u308b. \u70b9 \\((0,a)\\) \u3092\u901a\u308b \\(y=f(x)\\) \u306e\u30b0\u30e9\u30d5\u306e\u63a5\u7dda\u304c \\(1\\) \u672c\u3060\u3051\u5b58\u5728\u3059\u308b\u3088\u3046\u306a \\(a\\) \u306e\u5024\u3092\u6c42\u3081\u3088. \u307e\u305f, \\(a\\) \u304c\u305d\u306e\u5024\u3092\u3068\u308b\u3068\u304d, \\(y=f(x)\\) \u306e\u30b0\u30e9\u30d5, \u305d\u306e\u63a5\u7dda\u304a\u3088\u3073 \\(y\\) \u8ef8\u3067\u56f2\u307e\u308c\u305f\u56f3\u5f62\u306e\u9762\u7a4d\u3092\u6c42\u3081\u3088.<\/p><\/li>\r\n<\/ol>\r\n

    \r\n\r\n

    \u3010 \u89e3 \u7b54 \u3011<\/h4>\r\n

    (1)<\/strong><\/p>\r\n

    \\[\\begin{align}\r\nf'(x) & = (2x-1) e^{-x} -(x^2-x) e^{-x} \\\\\r\n& = -(x^2-3x+1) e^{-x} \\\\\r\n& = -\\left( x -\\dfrac{3-\\sqrt{5}}{2} \\right) \\left( x -\\dfrac{3+\\sqrt{5}}{2} \\right) e^{-x} , \\\\\r\nf''(x) & = -(2x-3) e^{-x} +(x^2-3x+1) e^{-x} \\\\\r\n& = (x^2-5x+4) e^{-x} \\\\\r\n& = (x-1)(x-4) e^{-x}\n\\end{align}\\]\r\n\u307e\u305f\r\n\\[\r\n\\displaystyle\\lim _ {x \\rightarrow -\\infty} f(x) = \\displaystyle\\lim _ {t \\rightarrow \\infty} (t^2+t) e^t = \\infty\n\\]\r\n\\(x \\gt 0\\) \u306b\u304a\u3044\u3066, \\(0 \\lt f(x) \\lt x^2 e^{-x}\\) \u3067\u3042\u308a\r\n\\[\r\n\\displaystyle\\lim _ {x \\rightarrow \\infty} x^2 e^{-x} = 0\n\\]\r\n\u306a\u306e\u3067, \u306f\u3055\u307f\u3046\u3061\u306e\u539f\u7406\u3088\u308a\r\n\\[\r\n\\displaystyle\\lim _ {x \\rightarrow \\infty} f(x) = 0\n\\]\r\n\u4ee5\u4e0a\u3088\u308a, \\(f(x)\\) \u306e\u5897\u6e1b\u3068\u51f9\u51f8\u306f\u4e0b\u8868\u306e\u3088\u3046\u306b\u306a\u308b.\r\n\\[\r\n\\begin{array}{c|ccccccccccc} x & ( -\\infty ) & \\cdots & \\frac{3-\\sqrt{5}}{2} & \\cdots & 1 & \\cdots & \\frac{3+\\sqrt{5}}{2} & \\cdots & 4 & \\cdots & ( \\infty ) \\\\ \\hline f'(x) & & - & 0 & + & & + & 0 & - & & - & \\\\ \\hline f''(x) & & + & & + & 0 & - & & - & 0 & + & \\\\ \\hline f(x) & ( \\infty ) & \\searrow ( \\cup ) & \\text{\u6975\u5c0f} & \\nearrow ( \\cup ) & 0 & \\nearrow ( \\cap ) & \\text{\u6975\u5927} & \\searrow ( \\cap ) & \\frac{12}{e^4} & \\searrow ( \\cup ) & (0) \\end{array}\r\n\\]\r\n\u3088\u3063\u3066, \\(y = f(x)\\) \u306e\u5909\u66f2\u70b9\u306f\r\n\\[\r\n\\underline{(1,0) , \\left( 4 , \\dfrac{12}{e^4} \\right)}\n\\]\r\n\u3067\u3042\u308a, \u30b0\u30e9\u30d5\u306e\u6982\u5f62\u306f\u4e0b\u56f3\u306e\u3088\u3046\u306b\u306a\u308b.<\/p>\r\n\"\"\r\n

    (2)<\/strong><\/p>\r\n

    \\(y = f(x)\\) \u306e \\(x\\) \u5ea7\u6a19\u304c \\(p\\) \u3067\u3042\u308b\u70b9\u306b\u304a\u3051\u308b\u63a5\u7dda\u306e\u65b9\u7a0b\u5f0f\u306f\r\n\\[\\begin{align}\r\ny & = (x-p) f'(p) +f(p) \\\\\r\n& = -(x-p)(p^2-3p+1) e^{-p} +(p^2-p) p^{-p} \\\\\r\n& = -(p^2-3p+1) e^{-p} x +p^2 (p-2) e^{-p}\n\\end{align}\\]\r\n\u3053\u308c\u304c\u70b9 \\((0,a) \\ ( a \\gt 0 )\\) \u3092\u901a\u308b\u3068\u304d\r\n\\[\r\na= p^2 (p-2) e^{-p} \\quad ... [1]\r\n\\]\r\n\u304c\u6210\u7acb\u3059\u308b.\r\n\u3057\u305f\u304c\u3063\u3066, [1] \u3092\u6e80\u305f\u3059\u6b63\u306e\u89e3 \\(p\\) \u304c\u305f\u3060 \\(1\\) \u3064\u5b58\u5728\u3059\u308b\u3088\u3046\u306a \\(a\\) \u306e\u5024\u3092\u6c42\u3081\u308c\u3070\u3088\u3044.
    \r\n[1] \u306e\u53f3\u8fba\u3092 \\(g(p)\\) \u3068\u304a\u304f.\r\n\\[\\begin{align}\r\ng'(p) & = (3p^2-4p) e^{-p} -(p^3-2p^2) e^{-p} \\\\\r\n& = -p (p^2-5p+4) e^{-p} \\\\\r\n& = -p (p-1)(p-4) e^{-p}\n\\end{align}\\]\r\n\u3057\u305f\u304c\u3063\u3066, \\(g(p)\\) \u306e\u5897\u6e1b\u306f\u4e0b\u8868\u306e\u3088\u3046\u306b\u306a\u308b.\r\n\\[\r\n\\begin{array}{c|ccccccccc} p & ( -\\infty ) & \\cdots & 0 & \\cdots & 1 & \\cdots & 4 & \\cdots & ( \\infty ) \\\\ \\hline g'(p) & & + & 0 & - & 0 & + & 0 & - & \\\\ \\hline g(p) & ( -\\infty ) & \\nearrow & 0 & \\searrow & -\\frac{1}{e} & \\nearrow & \\frac{32}{e^4} & \\searrow & (0) \\end{array}\r\n\\]\r\n\u306a\u306e\u3067 \\(y = g(p)\\) \u306e\u30b0\u30e9\u30d5\u306e\u6982\u5f62\u306f\u4e0b\u306e\u3088\u3046\u306b\u306a\u308b.<\/p>\r\n\"\"\r\n

    \u3053\u308c\u3068 \\(y = a\\) \u304c\u305f\u3060 \\(1\\) \u3064\u5171\u6709\u70b9\u3092\u3082\u3064\u306e\u306f\r\n\\[\r\na =\\underline{\\dfrac{32}{e^4}}\r\n\\]\r\n\u306e\u3068\u304d\u3067\u3042\u308b.
    \r\n\u3053\u306e\u3068\u304d, \\(y = f(x)\\) , \u63a5\u7dda, \\(y\\) \u8ef8\u304c\u56f2\u3080\u56f3\u5f62\u306f\u4e0b\u56f3\u659c\u7dda\u90e8\u3067\u3042\u308b.<\/p>\r\n\"\"\r\n

    \u63a5\u7dda\u306e\u65b9\u7a0b\u5f0f\u306f\r\n\\[\\begin{align}\r\ny & = -(4^2 -3 \\cdot 4 +1) e^{-4} x +32 e^{-4} \\\\\r\n& = -\\dfrac{5}{e^4} x +\\dfrac{32}{e^4}\n\\end{align}\\]\r\n\u3088\u3063\u3066, \u6c42\u3081\u308b\u9762\u7a4d \\(S\\) \u306f\r\n\\[\\begin{align}\r\nS & = \\displaystyle\\int _ 0^4 \\left\\{ -\\dfrac{5}{e^4} x +\\dfrac{32}{e^4} -(x^2-x) e^{-x} \\right\\} \\, dx \\\\\r\n& = -\\dfrac{5}{e^4} \\left[ \\dfrac{x^2}{2} \\right] _ 0^4 +\\dfrac{32}{e^4} \\cdot 4 +\\left[ (x^2-x) e^{-x} \\right] _ 0^4 -\\displaystyle\\int _ 0^4 (2x-1) e^{-x} \\, dx \\\\\r\n& = \\dfrac{88}{e^4} +\\dfrac{12}{e^4} +\\left[ (2x-1) e^{-x} \\right] _ 0^4 -\\displaystyle\\int _ 0^4 2 e^{-x} \\, dx \\\\\r\n& = \\dfrac{100}{e^4} +\\dfrac{7}{e^4} +1 +2 \\left[ e^{-x} \\right] _ 0^4 \\\\\r\n& = \\underline{\\dfrac{109}{e^4} -1}\n\\end{align}\\]\r\n","protected":false},"excerpt":{"rendered":"\u95a2\u6570 \\(f(x) = (x^2-x) e^{-x}\\) \u306b\u3064\u3044\u3066, \u6b21\u306e\u554f\u3044\u306b\u7b54\u3048\u3088. \u5fc5\u8981\u306a\u3089\u3070, \u4efb\u610f\u306e\u81ea\u7136\u6570 \\(n\\) \u306b\u5bfe\u3057\u3066 \\[ \\displaystyle\\lim _ {x \\rightarrow + … \u7d9a\u304d\u3092\u8aad\u3080 →<\/span><\/a>","protected":false},"author":1,"featured_media":0,"comment_status":"open","ping_status":"open","sticky":false,"template":"","format":"standard","meta":{"inline_featured_image":false,"footnotes":""},"categories":[34],"tags":[143,14],"class_list":["post-448","post","type-post","status-publish","format-standard","hentry","category-nagoya_r_2010","tag-nagoya_r","tag-14"],"_links":{"self":[{"href":"https:\/\/www.roundown.net\/nyushi\/wp-json\/wp\/v2\/posts\/448","targetHints":{"allow":["GET"]}}],"collection":[{"href":"https:\/\/www.roundown.net\/nyushi\/wp-json\/wp\/v2\/posts"}],"about":[{"href":"https:\/\/www.roundown.net\/nyushi\/wp-json\/wp\/v2\/types\/post"}],"author":[{"embeddable":true,"href":"https:\/\/www.roundown.net\/nyushi\/wp-json\/wp\/v2\/users\/1"}],"replies":[{"embeddable":true,"href":"https:\/\/www.roundown.net\/nyushi\/wp-json\/wp\/v2\/comments?post=448"}],"version-history":[{"count":0,"href":"https:\/\/www.roundown.net\/nyushi\/wp-json\/wp\/v2\/posts\/448\/revisions"}],"wp:attachment":[{"href":"https:\/\/www.roundown.net\/nyushi\/wp-json\/wp\/v2\/media?parent=448"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/www.roundown.net\/nyushi\/wp-json\/wp\/v2\/categories?post=448"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/www.roundown.net\/nyushi\/wp-json\/wp\/v2\/tags?post=448"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}