Question

求函数y=2x三次方+3x²-2x+10的极值和单调区间.急？

Answer 1

f(x)=2x^3+3x^2-2x+10
f'(x)=6x^2+6x-2
令f'(x)=0
6x^2+6x-2=0
x1=(-3+√21)/6
x2=(-3-√21)/6
f’’(x)=12x+6
f’’(x1)>0 f(x1)= 9.71788723 极小值
f’’(x2),2,ymax=y[x=-(7/12)^(1/2)-1/2],
ymin=y[x=(7/12)^(1/2)-1/2]
单增区间:(负无穷,-(7/12)^(1/2)-1/2)及((7/12)^(1/2)-1/2,正无穷),
单减区间(,-(7/12)^(1/2)-1/2,(7/12)^(1/2)-1/2),2,y=2x三次方+3x²-2x+10
y'=6x²+6x-2=0
即3x²+3x-1=0
解得：x1=（-3+根号21）/2
x2=（-3-根号21）/2
这就是极值,1,