设函数f(x)=X²+x-0.25（1）若定义域限制为[0,3],求f(x)的值域.
问题描述：
设函数f(x)=X²+x-0.25（1）若定义域限制为[0,3],求f(x)的值域.
（2）若定义域限制为[a,a+1]时,值域f(x)为[-0.5,0.0625],求a的值.
答
(1)f'(x)=(X²+x-0.25）'=2x+1
令f(x)=0,解得x=-1/2,-1/2不属于[0,3],所以f(x)在[0,3]上是单调的
f(x)的值域为[f(0),f(3)]即[-0.25,11.75]
(2)a^2+a-0.25=-0.5
a^2+a+0.25=0
(a+0.5)^2=0
解得a=-0.5
(a+1)^2+(a+1)-0.25=0.062
(a+1)^2+(a+1)+1/4=0.5625
(a+1+1/2)^2=0.5625
a+1+1/2=±0.75
a=-0.75,a=-2.25