Question

已知函数f(x)=ax²+(2a-1)x-3(a≠0),x∈[-3/2,2],求y=f(x)值域

Answer 1

f(x)=a[x+(2a-1)/(2a)]^2-(2a-1)^2/(4a)-3,x∈[-3/2,2],
a>0时
1）1/4<=(1-2a)/(2a)<=2,即1/6<=a<=2/5,
f(x)|min=-(2a-1)^2/(4a)-3=-(4a^2+8a+1)/(4a),
f(x)|max=f(-3/2)=9a/4-3a-3/2=-3a/4-3/2.
f(x)的值域是[-(4a^2+8a+1)/(4a),-3a/4-3/2];
2）-3/2<=(1-2a)/(2a)<1/4,即a>2/5,
f(x)|max=f(2)=8a-5,
f(x)的值域是[-(4a^2+8a+1)/(4a),8a-5];
3）(1-2a)/(2a)>2,即0<a<1/6时
f(x)|min=f(2),f(x)|max=f(-3/2),
∴f(x)的值域是[8a-5,-3a/4-3/2].
a<0时仿上讨论，留给您练习.

Answer 2

已知函数f(x)=ax^2+(2a-1)x-3(a≠0),x∈[-3/2,2],求y=f(x)值域
解析：当a>0时，函数f(x)为开口向上的抛物线，对称轴x=1/(2a)-1
令1/(2a)-1>=2==>a<=1/6
a∈(0,1/6],对称轴>=2，f(x)在[-3/2,2]上单调减；
f(-3/2)=9a/4-3(2a-1)/2-3=-3a/4-3/2
f(2)=4a+2(2a-1)-3=8a-5
∴y=f(x)值域为[8a-5,-(3a/4+3/2)]
令-3a/4-3/2=8a-5==>a=14/35=2/5
a∈(1/6,2/5],1/4<=对称轴<2
∴y=f(x)值域为[-(4a^+8a+1)/(4a),-(3a/4+3/2)]
a∈(2/5,+∞)，-1<对称轴<1/4
∴y=f(x)值域为[-(4a^+8a+1)/(4a),8a-5]

当a<0时，函数f(x)为开口向下的抛物线，对称轴x=1/(2a)-1
令1/(2a)-1<=-3/2==>a>=-1
a∈[-1,0),对称轴<=-3/2，f(x)在[-3/2,2]上单调减；
∴y=f(x)值域为[8a-5,-(3a/4+3/2)]
a∈(-∞,-1),-3/2<对称轴<-1
∴y=f(x)值域为[8a-5,-(4a^+8a+1)/(4a)]