设函数f(x)在x0处可导,且f'(x0)=-3,则曲线y=f(x)在点(x0,f(x0))处的切线的倾斜角为 设函数y=f(x)在x=x0处的导数为f'(x0),则曲线y...
\u8bbe\u51fd\u6570y=f\uff08x\uff09\u5728\u70b9x0\u5904\u6709\u5bfc\u6570\uff0c\u4e14f'(x0)\uff1e0\uff0c\u5219\u66f2\u7ebfy=f\uff08x\uff09\u5728\u70b9\uff08x0\uff0cf\uff08x0\uff09\uff09\u5904\u5207\u7ebf\u7684\u503e\u659c\u89d2\u7684\u8303\u56f4\u662f0\u5230\u03c0/2
\u6ca1\u4ec0\u4e48\u8fc7\u7a0b\u5427\uff0c\u4f5c\u4e2a\u89e3\u91ca\u597d\u4e86
\u7ebfy=f\uff08x\uff09\u5728\u70b9\uff08x0\uff0cf\uff08x0\uff09\uff09\u5904\u5207\u7ebf\u7684\u659c\u7387\u5373\u662f f'(x0)
\u659c\u7387\u5373\u662f \u503e\u659c\u89d2a\u7684\u6b63\u5207\u503c
\u5373 tan a = f'(x0) >0 \u6240\u4ee5\u3002\u3002\u4f60\u77e5\u9053\u7684\u3002
\u6ce8\uff1a\u6570\u5b66\u4e0a\u5207\u7ebf\u7684\u503e\u659c\u89d2\u7684\u8303\u56f4\u662f -\u03c0 \u5230 \u03c0
\u89e3\u7531\u51fd\u6570y=f(x)\u5728x=x0\u5904\u7684\u5bfc\u6570\u4e3af'(x0),
\u77e5\u51fd\u6570y=f\uff08x\uff09\u5728\u70b9\uff08x0\uff0cf\uff08x0\uff09\uff09\u5904\u7684\u5207\u7ebf\u7684\u659c\u7387k=f'(x0\uff09
\u53c8\u7531\u5207\u7ebf\u8fc7\u70b9\uff08x0\uff0cf\uff08x0\uff09\uff09
\u7531\u76f4\u7ebf\u70b9\u659c\u5f0f\u65b9\u7a0b\u77e5
\u5207\u7ebf\u65b9\u7a0b\u4e3ay-f\uff08x0\uff09=k\uff08x-x0\uff09
\u5373\u4e3ay-f\uff08x0\uff09=f'(x0\uff09\uff08x-x0\uff09
\u5373\u5207\u7ebf\u65b9\u7a0b\u4e3ay=f'(x0\uff09x-x0f'(x0\uff09+f\uff08x0\uff09\u3002
y=f(x)在x=x0处的导数为-3,也就是在x=x0处切线斜率为-3。
那么切线倾斜角是 arctan(-3)≈-71.5650512°
根据导数的几何意义：k=f'(x0)=-3
则tanθ=k=-3
∴θ=arctan(-3)=-arctan3
∵θ∈[0，π)
∴θ=π - arctan3
选择：C，因为根据导数的基本定义，可以解得
(Δf(x0)-df(x0))/Δx=df(x0)/dx-df(x0)/dx=0
绛旓細绠鍗曞垎鏋愪竴涓嬶紝绛旀濡傚浘鎵绀
绛旓細=lim x²f(x) / x^3 -2lim f(x³)/x³=lim f(x)/x -2lim f(x³)/x³=lim (f(x) - f(0) ) /(x-0) -2lim (f(x³)-f(0³))/(x³-0³)=f'(0)-2f'(0)= -f'(0)
绛旓細鍦ㄧ偣x0澶勫彲瀵硷紝鍒f(x)鍦鐐箈0鐨勬煇閭诲煙鍐呭繀瀹氳繛缁紝杩欏彞璇濇槸閿欒鐨勩備妇渚嬭鏄庯細f(x)=0锛屽綋x鏄湁鐞嗘暟 f(x)=x^2锛屽綋x鏄棤鐞嗘暟 鍙鍦▁=0澶勭偣杩炵画锛屽苟鍙锛屾寜瀹氫箟鍙獙璇佸湪x=0澶勫鏁颁负0 浣f(x) 鍦鍒殑鐐归兘涓嶈繛缁 鍑芥暟鍙鍒欏嚱鏁拌繛缁紱鍑芥暟杩炵画涓嶄竴瀹氬彲瀵硷紱涓嶈繛缁殑鍑芥暟涓瀹氫笉鍙銆
绛旓細鐢变簬鍒嗘瘝鏋侀檺涓0锛屽垯鍒嗗瓙鏋侀檺蹇呬负0锛屽洜姝im(x--->0)[f(x)+1]=0锛屽垯lim(x--->0)f(x)=-1銆傜敱f(x)鍦▁=0鍙锛鍒檉(x)鍦▁=1杩炵画锛屽洜姝鍑芥暟鍊间笌鏋侀檺鍊肩浉绛 f(0)=-1 lim [x--->0][f(x)+1]/(x+sinx)=lim [x--->0][f(x)-f(0)]/(x+sinx)=lim [x--->0][(...
绛旓細y-f(x0)=f鈥(x0)(x-x0)锛屽洜涓哄湪鐐x0澶鍙栧緱鏋佸皬鍊硷紝鎵浠鈥(x0)=0锛屽師寮忓寲涓簓-f(x0)=0锛寉=f(x0)銆傚畬姣曪紝鏈涢噰绾炽
绛旓細f'(x0)=lim[f(x0+鈻x)锛峟(x0)]/鈻硏 =lim[a(鈻硏)+b(鈻硏)^2]/鈻硏 =lim[a+b鈻硏]=a
绛旓細鏋侀檺lim锛堚柍x鈫0锛夛經銆恌锛坸0-鈻x锛銆/鈻硏锝 =f'(x0)鍑芥暟f(x0)鍦▁0澶勫彲瀵,涓f鈥橈紙x0锛=0鍚э紵閭ｆ牱灏辨槸 鏋侀檺lim锛堚柍x鈫0锛夛經銆恌锛坸0-鈻硏锛夈/鈻硏锝 =f'(x0)=0 鍚﹀垯涓嶈兘姹傚嚭鏉'(x0)鏄灏
绛旓細say shit to Baidu 鍨冨溇鐧惧害锛岃佸瓙缁冧範瀵兼暟棰橈紝鎯冲垹灏卞垹锛孋AO limx鈫0{[f(x)+1]/[x+sinx]}=2 (浠ヤ笅鐪佺暐x鈫0)[limf(x)+1]/limsinx=2 lim[f(x)+1]/lim2sinx=1 鍙f(x)+1鍜2sinx鏄瓑浠锋棤绌峰皬锛屽畠浠鍦▁->0鏃讹紝瓒嬭繎浜0鐨勯熷害鐩稿悓 鎵浠ュ綋x=0鏃讹紝[f(0)+1]'=[2sin0]'...
绛旓細鏄剧劧鏄敊鐨勶紝璇︽儏濡傚浘鎵绀
绛旓細璁惧嚱鏁癴锛坸锛夊湪x0澶勫彲瀵硷紝鍒 锛1锛夊嚱鏁拌繛缁紱锛2锛夊乏鏋侀檺=鍙虫瀬闄
