\u4ee4\u4e00\u4e2a\u51fd\u6570\u7684\u5b9a\u4e49\u57df\u4e3a $D,\\forall x\\in D,-x \\in D$ \u4e14 $f(x)$ \u4e3a\u5076\u51fd\u6570\u65f6\uff1a<\/p>\n
- \n
- $f(x)=f(-x)=f(|x|)$<\/li>\n
- \u51fd\u6570\u4e2d\u53ea\u80fd\u7531\u5076\u6b21\u9879\u5b58\u5728\u4f8b\u5982\uff1a$y=x^2$\u662f\u5076\u51fd\u6570\uff1b\u800c $y=(x^2+x)$ \u56e0\u4e3a\u6b21\u6570\u9879\u4e3a $1$ \u7684\u7cfb\u6570\u4e0d\u4e3a $0$ \u6545\u4e0d\u662f\u5076\u51fd\u6570<\/li>\n<\/ul>\n
\u5f53 $f(x)$ \u4e3a\u5947\u51fd\u6570\u65f6\uff1a<\/p>\n
- \n
- $-f(x)=f(-x)$<\/li>\n
- \u51fd\u6570\u4e2d\u53ea\u80fd\u6709\u5947\u6b21\u9879\u5b58\u5728<\/li>\n<\/ul>\n
\n\u77e5\u9053\u4e86\u4ee5\u4e0a\u7684\u57fa\u7840\u90e8\u5206\u4e4b\u540e\uff0c\u6211\u4eec\u5c31\u53ef\u4ee5\u505a\u8fdb\u4e00\u6b65\u7684\u63a2\u7a76\uff1a<\/p>\n
\u82e5 $f(x)$ \u5173\u4e8e $x=t$ \u5bf9\u79f0\uff0c\u5219\u6709\uff1a<\/p>\n
- \n
- $f(t-x)=f(t+x)$ <\/li>\n
- $f(2t-x)=f(x)$<\/li>\n
- $f(-x)= f(x+2t)$<\/li>\n<\/ul>\n
\u540c\u6837\u7684\uff0c\u82e5\u6211\u4eec\u77e5\u9053\u4e86\u4ee5\u4e0a\u4fe1\u606f\u4e2d\u7684\u4efb\u610f\u4e00\u6761\uff0c\u6211\u4eec\u4fbf\u53ef\u4ee5\u5f88\u5feb\u6377\u7684\u6c42\u51fa\u51fd\u6570\u7684\u5bf9\u79f0\u8f74\u3002<\/p>\n
\u82e5 $f(x)$ \u5173\u4e8e\u70b9 $(a,b)$ \u4e2d\u5fc3\u5bf9\u79f0\uff1a<\/p>\n
- \n
- $f(a-x)=-f(a+x)$ <\/li>\n
- $f(2a-x)=-f(x)$<\/li>\n
- $f(-x)= -f(x+2a)$<\/li>\n
- $f(a+t)+f(a-t)=b$<\/li>\n<\/ul>\n
\u5bf9\u52fe\u51fd\u6570<\/h1>\n
\u4ee4 $f(x)=x+\\frac{1}{x},x\\in (-\\infty, 0) \\cup (0,\\infty)$ \uff0c $f(x)$ \u662f\u4e00\u4e2a\u5947\u51fd\u6570\uff0c\u5982\u4e0b\u56fe\u6240\u793a\uff1a<\/p>\n
![\"piMxqPg.png\"](\"data:image\/svg+xml;base64,PCEtLUFyZ29uTG9hZGluZy0tPgo8c3ZnIHdpZHRoPSIxIiBoZWlnaHQ9IjEiIHhtbG5zPSJodHRwOi8vd3d3LnczLm9yZy8yMDAwL3N2ZyIgc3Ryb2tlPSIjZmZmZmZmMDAiPjxnPjwvZz4KPC9zdmc+\")
<\/div><\/p>\n\u503c\u57df\uff1a$(-\\infty,-2]\\cup [2,+\\infty]$<\/p>\n
\u5355\u8c03\u9012\u589e\u533a\u95f4\uff1a$(-\\infty, -1), (1, \\infty)$<\/p>\n
\u5355\u8c03\u9012\u51cf\u533a\u95f4\uff1a$(-1, 0), (0, 1)$<\/p>\n
\n\u4ee4 $f(x)=bx+\\frac{a}{x},x\\in (-\\infty, 0) \\cup (0,\\infty),a>0$ \u7531\u57fa\u672c\u4e0d\u7b49\u5f0f\u53ef\u5f97\uff1a$x+\\frac{a}{x}\\ge 2\\sqrt{a}$ \uff0c\u5728 $x = \\sqrt{a}$ \u7684\u65f6\u5019\u53ef\u4ee5\u53d6\u5230\u51fd\u6570\u5728\u7b2c\u4e00\u8c61\u9650\u5185\u7684\u6700\u5c0f\u503c\uff0c\u82e5 $x = -\\sqrt{a}$ \u65f6\uff0c\u53ef\u4ee5\u53d6\u5230\u51fd\u6570\u5728\u7b2c\u4e09\u8c61\u9650\u5185\u7684\u6700\u5927\u503c\u3002\u51fd\u6570\u4f1a\u8d8b\u8fd1\u4e8e $y=bx$<\/p>\n
\n\u4ee4 $f(x)=x-\\frac{1}{x},x\\in (-\\infty, 0) \\cup (0,\\infty)$ \uff0c $f(x)$ \u662f\u4e00\u4e2a\u5947\u51fd\u6570\uff0c\u5982\u4e0b\u56fe\u6240\u793a\uff1a<\/p>\n
<\/div><\/p>\n\u51fd\u6570\u7684\u51f9\u51f8\u6027<\/h1>\n
\u4ee4 $y=\\sqrt{x}\uff0c0<x_1<x_2$ \uff0c\u8bc1\u660e\uff1a$\\frac{f(x_1)+f(x_2)}{2} <f(\\frac{x_1+x_2}{2} )$ <\/p>\n
\u4ee3\u6570\u8bc1\u660e\uff1a<\/p>\n
$\\because \\frac{f(x_1)+f(x_2)}{2} = \\frac{\\sqrt{x_1}+\\sqrt{x_2}}{2}, f(\\frac{x_1+x_2}{2})=\\sqrt{\\frac{x_1+x_2}{2}}$<\/p>\n
\u4ee4 $\\sqrt{x_1} = a, \\sqrt{x_2} = b$<\/p>\n
$\\therefore x_1=a^2, x_2 = b^2$<\/p>\n
$\\therefore \\frac{a+b}{2}<\\sqrt{\\frac{a^2+b^2}{2}}$<\/p>\n
\u51e0\u4f55\u8bc1\u660e\uff1a<\/p>\n
\u5728 $y=\\sqrt{x}$ \u7684\u51fd\u6570\u56fe\u50cf\u4e0a\u4efb\u53d6\u4e24\u70b9 $A, B$<\/p>\n
$A(x_1, f(x_1)), B(x_2, f(x_2))$<\/p>\n
\u8fde\u63a5 $AB$ \u53d6\u7ebf\u6bb5 $AB$ \u7684\u4e2d\u70b9\u6807\u8bb0\u4e3a $N$ <\/p>\n
\u8fc7 $N$ \u4f5c $x$ \u8f74\u7684\u5782\u7ebf\u4ea4 $f(x)$ \u4e8e\u70b9 $M$<\/p>\n
$N(\\frac{x_1+x_2}{2}, \\frac{f(x_1)+f(x_2)}{2}), M(\\frac{x_1+x_2}{2}, f(\\frac{x_1+x_2}{2}))$<\/p>\n
\u70b9 $M$ \u5728 \u70b9 $N$ \u7684\u4e0a\u65b9\uff0c\u53ef\u4ee5\u8bc1\u660e $\\frac{f(x_1)+f(x_2)}{2} <f(\\frac{x_1+x_2}{2} )$ <\/p>\n
<\/div><\/p>\n","protected":false},"excerpt":{"rendered":"\u51fd\u6570\u7684\u56fe\u50cf\u53d8\u6362\u95ee\u9898 \u57fa\u7840\u90e8\u5206\uff1a $y=f(x)$ \u5de6\u79fb $n$ \u4e2a\u5355\u4f4d\u5f97\u5230\uff1a$y=f(x+n)$ $y=f(x […]<\/p>\n","protected":false},"author":1,"featured_media":0,"comment_status":"open","ping_status":"open","sticky":false,"template":"","format":"standard","meta":{"footnotes":""},"categories":[17],"tags":[26],"_links":{"self":[{"href":"http:\/\/ggapa.net:81\/wp-json\/wp\/v2\/posts\/805"}],"collection":[{"href":"http:\/\/ggapa.net:81\/wp-json\/wp\/v2\/posts"}],"about":[{"href":"http:\/\/ggapa.net:81\/wp-json\/wp\/v2\/types\/post"}],"author":[{"embeddable":true,"href":"http:\/\/ggapa.net:81\/wp-json\/wp\/v2\/users\/1"}],"replies":[{"embeddable":true,"href":"http:\/\/ggapa.net:81\/wp-json\/wp\/v2\/comments?post=805"}],"version-history":[{"count":1,"href":"http:\/\/ggapa.net:81\/wp-json\/wp\/v2\/posts\/805\/revisions"}],"predecessor-version":[{"id":806,"href":"http:\/\/ggapa.net:81\/wp-json\/wp\/v2\/posts\/805\/revisions\/806"}],"wp:attachment":[{"href":"http:\/\/ggapa.net:81\/wp-json\/wp\/v2\/media?parent=805"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"http:\/\/ggapa.net:81\/wp-json\/wp\/v2\/categories?post=805"},{"taxonomy":"post_tag","embeddable":true,"href":"http:\/\/ggapa.net:81\/wp-json\/wp\/v2\/tags?post=805"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}
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<\/div><\/p>\n![](\"https:\/\/z1.ax1x.com\/2023\/11\/04\/piMzlJe.png\")
<\/div><\/p>\n![\"\"](\"https:\/\/z1.ax1x.com\/2023\/11\/04\/piQSp6A.png\")
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