设函数f(x)=x²+ln(2x+3);求f(x)在区间[-3/4,1/4]的值域

f'(x)=2x+2/(2x+3)=(4x^2+6x+2)/(2x+3)=2(x+1)(2x+1)/(2x+3)
[-3/4,-1) -1 (-1,-1/2) -1/2 (-1/2,1/4)
x+1 - 0 + + +
2x+1 - - - 0 +
2x+3 + + + + +
f'(x) + 0 - 0 +
f(x) 增 大 减 小 增
maxf(-1)=1 minf(-1/2)=1/4+ln2=0.9431
f(-3/4)=9/16+ln(3/2)=0.9680 f(1/4)=1/16+ln(7/2)=1.3153
f(x)在区间[-3/4,1/4]的值域:[1/4+ln2, 1/16+ln(7/2)]