Question: On the graph of f(x)=cosx and the domain 2\pi <=x<=4\pi , for which of the following intervals is f(x) strictly increasing? Choose the lower and the upper bound of this interval. On the graph of f ( x ) = cos x and the domain 2 ...
(π/2) = 0 sub into y=-2x + c, we get c = π/2So, tangent line is y = -2x + π/2Ans : None of the above(typo happened with your ans)2.d/ dxcos^2(x^3)=d[cos^2x^3]/dx=-2(cosx^3)(sinx^3)(3x^2)=-3x^2sin^2(2x^3) 3.0.5e.5 dt= 0.5(e.5)t + c (...
So,tangent line is y = -2x + π/2Ans :None of the above (typo happened with your ans) 2.d/dx cos^2(x^3)=d[cos²x³]/dx = -2(cosx³)(sinx³)(3x²) = -3x²sin²(2x³)3.0.5∫e^0.5 dt= 0.5(e^0.5)t + c (maybe mistyping)0.5∫e^(0.5t) dt= (e^0.5...
数学函数图像为您作cosx^2 cosx (cosx)^2的函数图像。
y=cosx/(2+sinx) y‘=[-sinx(2+sinx)-(cosx)^2]/(2+sinx)^2 =-(2sinx+1)/(2+sinx)^2 可知y’=0的点的切线是水平的 所以2sinx=-1 x=7π/6或11π/6 所以点(7π/6,1/3)(11π/6,-1/3)出的切线是水平的 分析总结。 在02pie之间哪些点使得函数的切线是垂直的结果...
Equations Unfold Math OP Unfold Linear algebra Unfold Derivative function Function image Hot issues Function image: The function image taken from the image library. Current location:Function image > Function Image Library > Image返回豫ICP备19044667号 ...
Equations Unfold Math OP Unfold Linear algebra Unfold Derivative function Function image Hot issues Function image: The function image taken from the image library. Current location:Function image > Function Image Library > Image返回豫ICP备19044667号 ...
Equations Unfold Math OP Unfold Linear algebra Unfold Derivative function Function image Hot issues Function image: The function image taken from the image library. Current location:Function image > Function Image Library > Image返回豫ICP备19044667号 ...
首先把极坐标方程转化为直角坐标方程x=r×cosxy=r×sinx 这里的r是极距 x是夹角 (一般夹角用θ表示)得cosx=x/r sinx=y/r 带入极坐标方程r=1/(x/r+y/r) 化简得到直角坐标方程为 x+y=1也就是直线 y=1-x与两个坐标轴围成的面积 答案是1/2...
(π/2) = 0 sub into y = -2x + c,we get c = π/2So,tangent line is y = -2x + π/2Ans :None of the above (typo happened with your ans) 2.d/dx cos^2(x^3)=d[cos²x³]/dx = -2(cosx³)(sinx³)(3x²) = -3x²sin²(2x³)3.0.5∫e^0.5 dt= 0.5(e...