y=sin(x 1)在x=-1处的切线方程为-搜答案-搜搜做题作业网

求y=sin(x+1)周期

y=Asin（ωx+ψ）周期为：T=2π/ωy=sin(x+1),ω=1,所以T=2π

由题意,f(x)=a(x-x1)(x-x2)(x-x3)则f'(x)=a(x-x2)(x-x3)+a(x-x1)(x-x3)+a(x-x1)(x-x2)令S=a(x1-x2)(x1-x3)(x2-x3

是奇函数,还是震荡函数,函数值在[-1,1]之间震荡,在x=0处没有极限

sin(x+y)sin(x-y)=-1/2(cos(x+y+x-y)—cos(x+y-x+y))=-1/2(cos2x—cos2y)=-1/2(1-2（sinx）^2-1+2(siny）^2)=（si

x1*x2=a-1x1+x2=-（a-2）因为点P(x1,x2)在圆x²+y²=4上所以x1²+x2²=4即（x1+x2）²-2x1*x2=4所以（a

由题意可知f(x1)=f(x)min=-1=>sin(π/2x1+π/3)=-1=>π/2x1+π/3=2k1π-π/2=>x1=1/(4k1-5/3)同理f(x2)=f(x)max=1=>sin(π

x=-pi:0.001:pi;y=sin(x)+1./sin(x);plot(x,y,'r',y,x,'b')矩阵元素运算需要加“.”

symsxyeq=cos(x*y)*cos(x*(1-y))-0.5*x*sin(x*y)*sin(x*(1-y))-1;ezplot(eq)

sysxabf1=x+1;f2=0.5*x^2;int(f1,0,1)+int(f2,1,2)f=exp(ax)*sin(bx)inf(f)

Limit[x^2Sin[1/x],x->0]=0;Limit[2xSin[1/x]-Cos[1/x],x->0]确实没有极限.函数y的定义域不是全体实数,即函数是间断的,极限就可能不存在!仔细考虑一

设x1

y=sin-1 x 求导

dx/dt={1/[2√(1-t^2)]}(-2t)=-t/√(1-t^2)dy/dt=1/√(1-t^2)dy/dx=[1/√(1-t^2)]/[-t/√(1-t^2)]=-1/t再问：为啥dy/d

由题意可知f(x1)=f(x)min=-1=>sin(π/2x1+π/3)=-1=>π/2x1+π/3=2k1π-π/2=>x1=1/(4k1-5/3)同理f(x2)=f(x)max=1=>sin(π

y=1-sin x如何作图?

先画y＝sinx.y＝－sinx关于x轴对称,再画y＝－sinx＋1,将图像向上平移一个单位就可以了.

x=0:0.1:100;%假设步长为0.1y=x.*sin(x);ind_peak=intersect(find(diff(y)>0)+1,find(diff(y)

对这样的隐函数求导数的时候,就把y看作x的函数,y对x求导就得到dy/dx所以原等式对x求导得到2xy²+x²*2y*dy/dx+siny+x*cosy*dy/dx=0于是化简得到

因为sin最值是-1和1所以f(x1)

x=pi:pi/50:4*pi;y=x.*tan(1./x).*sin(x.^3);plot(x,y),gridon>>程序给你改了下,看看,我执行时就没错了,也画出图了!

求导：y=3^(sin*1/x)

y=3^[sin(1/x)]y'=3^[sin(1/x)]ln3*cos(1/x)*(-1/x^2)=-ln3*3^[sin(1/x)]*cos(1/x)/x^2

f[(x1+x2)/2]=2^[(x1+x2)/2][f(x1)+f(x2)]/2=(2^x1+2^x2)/2由基本不等式(2^x1+2^x2)/2≧√[(2^x1)(2^x2)]=2^[(x1+x2