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

问题描述:

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...