来自苏岩的问题
函数f(x)=ex+x2-2在区间(-2,1)内零点的个数为()A.1B.2C.3D.4
函数f(x)=ex+x2-2在区间(-2,1)内零点的个数为()
A.1
B.2
C.3
D.4
1回答
2020-04-25 01:41
函数f(x)=ex+x2-2在区间(-2,1)内零点的个数为()A.1B.2C.3D.4
函数f(x)=ex+x2-2在区间(-2,1)内零点的个数为()
A.1
B.2
C.3
D.4
∵f(x)=ex+x2-2得f'(x)=ex+2xf''(x)=ex+2>0从而f'(x)是增函数,f'(-2)=1e2-4<0f'(0)=1>0从而f'(x)在(-2,1)内有唯一零点x0,满足则在区间(-2,x0)上,有f'(x)<0,f(x)是减函数,在区间(x...