解答
f(x)=−2sin(x)−4sin(2x)
解答
Period:2π
Domain:−∞<x<∞
Range:−21−12865−129−4sin(2arccos(16129−1))≤f(x)≤21−12865−129−4sin(2(2π−arccos(16129−1)))
X截距:(2πn,0),(π+2πn,0),(arccos(−41)+2πn,0),(−arccos(−41)+2π+2πn,0),Y截距:(0,0)
Asymptotes:无
ExtremePoints:极小值arccos(16129−1)+2πn,−21−12865−129−4sin(2arccos(16129−1)),极大值arccos(−161+129)+2πn,−21−12865+129−4sin(2arccos(−161+129)),极小值−arccos(−161+129)+2π+2πn,21−12865+129−4sin(2(−arccos(−161+129)+2π)),极大值2π−arccos(16129−1)+2πn,21−12865−129−4sin(2(2π−arccos(16129−1)))
+1
间隔符号
Domain:(−∞,∞)
Range:−21−12865−129−4sin(2arccos(16129−1)),21−12865−129−4sin(2(2π−arccos(16129−1)))
求解步骤
−2sin(x)−4sin(2x)的周期:2π
−2sin(x)−4sin(2x)的定义域 :−∞<x<∞
−2sin(x)−4sin(2x)的值域:−21−12865−129−4sin(2arccos(16129−1))≤f(x)≤21−12865−129−4sin(2(2π−arccos(16129−1)))
−2sin(x)−4sin(2x)的轴截距点:X 截距:(2πn,0),(π+2πn,0),(arccos(−41)+2πn,0),(−arccos(−41)+2π+2πn,0),Y 截距:(0,0)
−2sin(x)−4sin(2x)的渐近线:无
−2sin(x)−4sin(2x)的极值点:极小值arccos(16129−1)+2πn,−21−12865−129−4sin(2arccos(16129−1)),极大值arccos(−161+129)+2πn,−21−12865+129−4sin(2arccos(−161+129)),极小值−arccos(−161+129)+2π+2πn,21−12865+129−4sin(2(−arccos(−161+129)+2π)),极大值2π−arccos(16129−1)+2πn,21−12865−129−4sin(2(2π−arccos(16129−1)))