\u4e00\u822c\u5730\uff0c\u5982\u679c$x$\u4e0e$y$\u5173\u4e8e\u67d0\u79cd\u5bf9\u5e94\u5173\u7cfb <\/p>\n \u6027\u8d28\uff1a \u57fa\u672c\u521d\u7b49\u51fd\u6570\u5305\u62ec\u5e42\u51fd\u6570\u3001\u6307\u6570\u51fd\u6570\u3001\u5bf9\u6570\u51fd\u6570\u3001\u4e09\u89d2\u51fd\u6570\u3001\u53cd\u4e09\u89d2\u51fd\u6570\u548c\u5e38\u6570\u51fd\u6570\u3002<\/p>\n \u4e00\u822c\u5730\uff0c\u5f62\u5982 \u4f8b\u5982\u51fd\u6570 \u6307\u6570\u51fd\u6570\u4e2d\uff0c\u524d\u9762\u7684\u7cfb\u6570\u4e3a1\u3002\u5982\uff1a \u5f53 \u6307\u6570\u51fd\u6570\u8fd0\u7b97\u6cd5\u5219\uff1a<\/p>\n \u5bf9\u6570\u51fd\u6570\u662f\u6307\u6570\u51fd\u6570\u7684\u53cd\u51fd\u6570<\/p>\n \u5f53 \u5bf9\u6570\u51fd\u6570\u8fd0\u7b97\u6cd5\u5219\uff1a<\/p>\n \u6b63\u5f26\uff1a \u4e09\u89d2\u51fd\u6570\u516c\u5f0f<\/p>\n \uff081\uff09\u548c\u5dee\u89d2\u516c\u5f0f<\/p>\n \uff082\uff09\u548c\u5dee\u5316\u79ef\u516c\u5f0f<\/p>\n \uff083\uff09\u79ef\u5316\u548c\u5dee\u516c\u5f0f<\/p>\n \uff084\uff09\u4e8c\u500d\u89d2\u516c\u5f0f<\/p>\n \u53cd\u51fd\u6570 \u4e00\u822c\u5730\uff0c\u5982\u679c$x$\u4e0e$y$\u5173\u4e8e\u67d0\u79cd\u5bf9\u5e94\u5173\u7cfb$ f(x) $\u76f8\u5bf9\u5e94\uff0c$ y=f(x) $\uff0c\u5219$ y=f( […]<\/p>\n","protected":false},"author":1,"featured_media":0,"comment_status":"open","ping_status":"open","sticky":false,"template":"","format":"standard","meta":{"footnotes":""},"categories":[509],"tags":[],"class_list":["post-2117","post","type-post","status-publish","format-standard","hentry","category-mathematics-fundamentals"],"_links":{"self":[{"href":"https:\/\/www.appblog.cn\/index.php\/wp-json\/wp\/v2\/posts\/2117","targetHints":{"allow":["GET"]}}],"collection":[{"href":"https:\/\/www.appblog.cn\/index.php\/wp-json\/wp\/v2\/posts"}],"about":[{"href":"https:\/\/www.appblog.cn\/index.php\/wp-json\/wp\/v2\/types\/post"}],"author":[{"embeddable":true,"href":"https:\/\/www.appblog.cn\/index.php\/wp-json\/wp\/v2\/users\/1"}],"replies":[{"embeddable":true,"href":"https:\/\/www.appblog.cn\/index.php\/wp-json\/wp\/v2\/comments?post=2117"}],"version-history":[{"count":0,"href":"https:\/\/www.appblog.cn\/index.php\/wp-json\/wp\/v2\/posts\/2117\/revisions"}],"wp:attachment":[{"href":"https:\/\/www.appblog.cn\/index.php\/wp-json\/wp\/v2\/media?parent=2117"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/www.appblog.cn\/index.php\/wp-json\/wp\/v2\/categories?post=2117"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/www.appblog.cn\/index.php\/wp-json\/wp\/v2\/tags?post=2117"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}$ f(x) $<\/code>\u76f8\u5bf9\u5e94\uff0c$ y=f(x) $<\/code>\uff0c\u5219$ y=f(x) $<\/code>\u7684\u53cd\u51fd\u6570\u4e3a$ x=f(y) $<\/code>\u6216\u8005$ y=f^{-1}(x) $<\/code>\u3002
\n\u5b58\u5728\u53cd\u51fd\u6570(\u9ed8\u8ba4\u4e3a\u5355\u503c\u51fd\u6570\uff09\u7684\u6761\u4ef6\u662f\u539f\u51fd\u6570\u5fc5\u987b\u662f\u4e00\u4e00\u5bf9\u5e94\u7684\uff08\u4e0d\u4e00\u5b9a\u662f\u6574\u4e2a\u6570\u57df\u5185\u7684\uff09\u3002
\n\u6ce8\u610f\uff1a\u4e0a\u6807"\u22121"\u6307\u7684\u5e76\u4e0d\u662f\u5e42\u3002\u6700\u5177\u6709\u4ee3\u8868\u6027\u7684\u53cd\u51fd\u6570\u5c31\u662f\u5bf9\u6570\u51fd\u6570\u4e0e\u6307\u6570\u51fd\u6570\u3002<\/p>\n
\n\uff081\uff09\u51fd\u6570$ f(x) $<\/code>\u4e0e\u5b83\u7684\u53cd\u51fd\u6570$ f^{-1}(x) $<\/code>\u56fe\u8c61\u5173\u4e8e\u76f4\u7ebf$y=x$\u5bf9\u79f0\uff1b
\n\uff082\uff09\u51fd\u6570\u5b58\u5728\u53cd\u51fd\u6570\u7684\u5145\u8981\u6761\u4ef6\u662f\uff0c\u51fd\u6570\u7684\u5b9a\u4e49\u57df\u4e0e\u503c\u57df\u662f\u4e00\u4e00\u6620\u5c04\uff1b
\n\uff083\uff09\u4e00\u4e2a\u51fd\u6570\u4e0e\u5b83\u7684\u53cd\u51fd\u6570\u5728\u76f8\u5e94\u533a\u95f4\u4e0a\u5355\u8c03\u6027\u4e00\u81f4\uff1b
\n\uff084\uff09\u5927\u90e8\u5206\u5076\u51fd\u6570\u4e0d\u5b58\u5728\u53cd\u51fd\u6570\uff08\u5f53\u51fd\u6570$ y=f(x) $<\/code>\uff0c \u5b9a\u4e49\u57df\u662f$ \\\\{0\\\\} $<\/code> \u4e14$ f(x)=C $<\/code>\uff08\u5176\u4e2d$C$\u662f\u5e38\u6570\uff09\uff0c\u5219\u51fd\u6570$ f(x) $<\/code>\u662f\u5076\u51fd\u6570\u4e14\u6709\u53cd\u51fd\u6570\uff0c\u5176\u53cd\u51fd\u6570\u7684\u5b9a\u4e49\u57df\u662f$ \\\\{C\\\\} $<\/code>\uff0c\u503c\u57df\u4e3a$ \\\\{0\\\\} $<\/code>\uff09\u3002\u5947\u51fd\u6570\u4e0d\u4e00\u5b9a\u5b58\u5728\u53cd\u51fd\u6570\uff0c\u88ab\u4e0e$ y $<\/code>\u8f74\u5782\u76f4\u7684\u76f4\u7ebf\u622a\u65f6\u80fd\u8fc72\u4e2a\u53ca\u4ee5\u4e0a\u70b9\u5373\u6ca1\u6709\u53cd\u51fd\u6570\u3002\u82e5\u4e00\u4e2a\u5947\u51fd\u6570\u5b58\u5728\u53cd\u51fd\u6570\uff0c\u5219\u5b83\u7684\u53cd\u51fd\u6570\u4e5f\u662f\u5947\u51fd\u6570\u3002
\n\uff085\uff09\u4e00\u6bb5\u8fde\u7eed\u7684\u51fd\u6570\u7684\u5355\u8c03\u6027\u5728\u5bf9\u5e94\u533a\u95f4\u5185\u5177\u6709\u4e00\u81f4\u6027\uff1b
\n\uff086\uff09\u4e25\u589e\uff08\u51cf\uff09\u7684\u51fd\u6570\u4e00\u5b9a\u6709\u4e25\u683c\u589e\uff08\u51cf\uff09\u7684\u53cd\u51fd\u6570\uff1b
\n\uff087\uff09\u53cd\u51fd\u6570\u662f\u76f8\u4e92\u7684\u4e14\u5177\u6709\u552f\u4e00\u6027\uff1b
\n\uff088\uff09\u5b9a\u4e49\u57df\u3001\u503c\u57df\u76f8\u53cd\u5bf9\u5e94\u6cd5\u5219\u4e92\u9006\u3002<\/p>\n\u521d\u7b49\u51fd\u6570<\/h2>\n
\u5e42\u51fd\u6570<\/h3>\n
$ y=x^\u03b1 $<\/code>($ \u03b1 $<\/code>\u4e3a\u6709\u7406\u6570\uff09\u7684\u51fd\u6570\uff0c\u5373\u4ee5\u5e95\u6570\u4e3a\u81ea\u53d8\u91cf\uff0c\u5e42\u4e3a\u56e0\u53d8\u91cf\uff0c\u6307\u6570\u4e3a\u5e38\u6570\u7684\u51fd\u6570\u79f0\u4e3a\u5e42\u51fd\u6570\u3002
\n\u5176\u4e2d\uff0c$ \u03b1 $<\/code>\u53ef\u4e3a\u4efb\u4f55\u5e38\u6570\uff0c\u4f46\u4e2d\u5b66\u9636\u6bb5\u4ec5\u7814\u7a76$\u03b1$\u4e3a\u6709\u7406\u6570\u7684\u60c5\u5f62\uff08$\u03b1$\u4e3a\u65e0\u7406\u6570\u65f6\u53d6\u5176\u8fd1\u4f3c\u7684\u6709\u7406\u6570\uff09<\/p>\n$ y=C$\u3001$y=x$\u3001$y=x^2$\u3001$y=x^{-1}=\\frac{1}{x} (x\u22600) $<\/code>\u3001$ y=x^{\\frac{1}{2}}=\\sqrt{x} $<\/code>\u7b49\u90fd\u662f\u5e42\u51fd\u6570\u3002<\/p>\n\u6307\u6570\u51fd\u6570<\/h3>\n
$y=a^x $<\/code><\/p>\n$ a $<\/code>\u4e3a\u5e38\u6570\u4e14$ a>0\uff0ca\u22601 $<\/code>\u79f0\u4e3a\u6307\u6570\u51fd\u6570\uff0c\u5176\u4e2d$ x \\in R$\uff0c$y \\in (0\uff0c+\u221e) $<\/code>\u3002<\/p>\n$ y=10^x $<\/code>\u3001$ y=\\pi^x $<\/code> \u90fd\u662f\u6307\u6570\u51fd\u6570\uff0c\u800c$ y=2*3^x $<\/code>\u4e0d\u662f\u6307\u6570\u51fd\u6570<\/p>\n$ a>1 $<\/code>\u65f6\uff0c\u6307\u6570\u51fd\u6570\u662f\u5355\u8c03\u9012\u589e\u51fd\u6570\uff1b\u5f53$ 0<a<1 $<\/code>\u65f6\uff0c\u6307\u6570\u51fd\u6570\u662f\u5355\u8c03\u9012\u51cf\u51fd\u6570\u3002<\/p>\n$ a^{\\frac{1}{n}}=\\sqrt[n]{a} $<\/code>
\n$ a^{m+n}=a^m\u00b7a^n $<\/code>
\n$ a^{m-n}=a^m\u00b7a^{-n}=\\frac{a^m}{a^n} $<\/code>
\n$ a^{mn}=(a^m)^n=(a^n)^m $<\/code><\/p>\n\u5bf9\u6570\u51fd\u6570<\/h3>\n
$ y=\\log_ax $<\/code><\/p>\n$ a $<\/code>\u4e3a\u5e38\u6570\u4e14$a>0\uff0ca\u22601$\u79f0\u4e3a\u5bf9\u6570\u51fd\u6570\uff0c$ x \\in (0\uff0c+\u221e) $<\/code>\uff0c$ y \\in R $`\u3002<\/p>\n$ a>1 $<\/code>\u65f6\uff0c\u6307\u6570\u51fd\u6570\u662f\u5355\u8c03\u9012\u589e\u51fd\u6570\uff1b\u5f53$ 0<a<1 $<\/code>\u65f6\uff0c\u6307\u6570\u51fd\u6570\u662f\u5355\u8c03\u9012\u51cf\u51fd\u6570\u3002<\/p>\n$ \\log_aMN=\\log_aM+\\log_aN $<\/code>
\n$ \\log_a\\frac{M}{N}=\\log_aM-\\log_aN $<\/code>
\n$ \\log_aa^b=b $<\/code>, $ \\log_{10}a=\\operatorname{lg}a $<\/code>, $ \\log_ea=\\operatorname{ln}a $<\/code>
\n$ \\log_ab=\\frac{\\log_cb}{\\log_ca} $<\/code> \/\/\u6362\u5e95\u516c\u5f0f
\n$ \\log_{a^n}b^m=\\frac{m}{n}log_ab $<\/code><\/p>\n\u4e09\u89d2\u51fd\u6570<\/h3>\n
$ y=\\operatorname{sin}x$\uff0c$y \\in [-1\uff0c1] $<\/code>
\n\u4f59\u5f26\uff1a$ y=\\operatorname{cos}x$\uff0c$y \\in [-1\uff0c1] $<\/code>
\n\u6b63\u5207\uff1a$ y=\\operatorname{tan}x$\uff0c$y \\in (-\u221e\uff0c+\u221e) $<\/code>
\n\u6b63\u5207\uff1a$ y=\\operatorname{cot}x$\uff0c$y \\in (-\u221e\uff0c+\u221e) $<\/code><\/p>\n$ \\operatorname{cos}(a+b)=\\operatorname{cos}a \\operatorname{cos}b - \\operatorname{sin}a \\operatorname{sin}b $<\/code>
\n$ \\operatorname{cos}(a-b)=\\operatorname{cos}a \\operatorname{cos}b + \\operatorname{sin}a \\operatorname{sin}b $<\/code>
\n$ \\operatorname{sin}(a+b)=\\operatorname{sin}a \\operatorname{cos}b + \\operatorname{cos}a \\operatorname{sin}b $<\/code>
\n$ \\operatorname{sin}(a-b)=\\operatorname{sin}a \\operatorname{cos}b - \\operatorname{cos}a \\operatorname{sin}b $<\/code>
\n$ \\operatorname{tan}(a+b)=\\frac{\\operatorname{tan}a + \\operatorname{tan}b}{1-\\operatorname{tan}a \\operatorname{tan}b} $<\/code>
\n$ \\operatorname{tan}(a-b)=\\frac{\\operatorname{tan}a - \\operatorname{tan}b}{1+\\operatorname{tan}a \\operatorname{tan}b} $<\/code><\/p>\n$ \\operatorname{sin}a+\\operatorname{sin}b=2\\operatorname{sin}\\frac{a+b}{2}\\operatorname{cos}\\frac{a-b}{2} $<\/code>
\n$ \\operatorname{sin}a-\\operatorname{sin}b=2\\operatorname{cos}\\frac{a+b}{2}\\operatorname{sin}\\frac{a-b}{2} $<\/code>
\n$ \\operatorname{cos}a+\\operatorname{cos}b=2\\operatorname{cos}\\frac{a+b}{2}\\operatorname{cos}\\frac{a-b}{2} $<\/code>
\n$ \\operatorname{cos}a-\\operatorname{cos}b=-2\\operatorname{sin}\\frac{a+b}{2}\\operatorname{sin}\\frac{a-b}{2} $<\/code>
\n$ \\operatorname{tan}a\u00b1\\operatorname{tan}b=\\frac{\\operatorname{sin}(a\u00b1b)}{\\operatorname{cos}a\\operatorname{cos}b} $<\/code>
\n$ \\operatorname{cot}a\u00b1\\operatorname{cot}b=\u00b1\\frac{\\operatorname{sin}(a\u00b1b)}{\\operatorname{sin}a\\operatorname{sin}b} $<\/code><\/p>\n$ \\operatorname{sin}a\\operatorname{cos}b=\\frac{1}{2}[\\operatorname{sin}(a+b)+\\operatorname{sin}(a-b)] $<\/code>
\n$ \\operatorname{cos}a\\operatorname{sin}b=\\frac{1}{2}[\\operatorname{sin}(a+b)-\\operatorname{sin}(a-b)] $<\/code>
\n$ \\operatorname{cos}a\\operatorname{cos}b=\\frac{1}{2}[\\operatorname{cos}(a+b)+\\operatorname{cos}(a-b)] $<\/code>
\n$ \\operatorname{sin}a\\operatorname{sin}b=\\frac{1}{2}[\\operatorname{cos}(a+b)-\\operatorname{cos}(a-b)] $<\/code><\/p>\n$ \\operatorname{sin}2a=2\\operatorname{sin}a\\operatorname{cos}a $<\/code>
\n$ \\operatorname{cos}2a=2\\operatorname{cos}^2a-1=1-2\\operatorname{sin}^2a $<\/code>
\n$ \\operatorname{tan}2a=\\frac{2\\operatorname{tan}a}{1-\\operatorname{tan}^2a} $<\/code><\/p>\n","protected":false},"excerpt":{"rendered":"