y=x-x三次方的单调增区间为-搜答案-智慧作业帮

f(x)导数=3x^2-6x令导数等于0x=0或2(-∞,0)(2,+∞)单增(0,2)单减

y=x^3+mx^2+nx+qy'=3x^2+2mx+nx=-1,3*(-1)^2+2m*(-1)+n=0,3-2m+n=0.(1)x=2,3*2^2+2m*2+n=0,12+4m+n=0.(2)(2

y'=6x²+6x=0x=-1,x=0x0,y'>0,递增-1

y'=6x^2-6x令y'=0得x=0,1在(-∞,0),(1,+∞)内,y'＞0,y单调上升在(0,1)内,y'

y'=6x^2+6x-12>0x>1或x

y'=3x^2-3=3(x-1)(x+1)=0,得极值点x=-1,1单调增区间：（-∞,-1）,（1,+∞）单调减区间：（-1,1）极大值y(-1)=-1+3-2=0极小值y(1)=1-3-2=-4

y'=-3x^2

f'(x)=3x^2-12=3(x+2)(x-2)令f'>0,得x2,这时f(x)单增令f'

定义域为x≠0y'=(x³)'+(3/x)'=3x²-3/x²>0x²-1/x²>0所以(x²)²-1>0所以x²>1所

y=x³+3x²-1y′=3x²+6x当y′=0即3x²+6x=03x(x+2)=0x=0x=-2当x=0函数单调增当-2

f(x)=2x^3+3x^2-2x+10f'(x)=6x^2+6x-2令f'(x)=06x^2+6x-2=0x1=(-3+√21)/6x2=(-3-√21)/6f’’(x)=12x+6f’’(x1)>

f（x）=x^3-3x^2+2f`(x)=3x^2-6x

解由y=x-x^3求导得f'(x)=1-3x^2令f'(x)=0解得x=±√3/3.故当x属于（负无穷大,-√3/3)时,f'(x)＜0当x属于（-√3/3,√3/3)时,f'(x)＞0当x属于（-√

y=2^(-3+4x-x^2)=2^[-(x-2)^2+1)]当x≤2时,y随x的增大也即单增区间为(-∞,2],单减区间为(2,+∞],函数值域为(1,2]..

y=2^(-x²-3x+2)-x²-3x+2的对称轴是x=-3/2求函数y=2^(-x²-3x+2)的单调增区间就是求函数y=-x²-3x+2的单调增区间因为y

y=x²-x³y'=2x-3x²=x(2-3x)得极值点为x=0,2/3单调增区间为(0,2/3)单调减区间为(-∞,0),(2/3,+∞)再问：极值点是怎么求得呀~再答

微积分学过没有?学过就好说了：f(x)=x^3+3x^2-9x+16,f'(x)=3x^2+6x-9=3(x-1)(x+3)当x=1或-3时,f'(x)=0,由(x-1)(x+3)>0,可解得：x1由

f(x)=x^3/3+x^2-3x一阶导数为：f'(x)=x^2+2x-3图像在原点处的切线斜率为：f'(0)=0+0-3=-3则图像在原点处的切线方程y=-3x当f'(x)