Question

已知函数f(x)=ax2-|x-a|.

当a＞0时，求函数f(x)在区间[3，＋∞)上的值域．

Answer 1

f(x)=ax^2-|x-a|
①a<=3时,f(x)=ax^2-(x-a)=ax^2-x+a
X(对称轴)=1/2a
Ⅰ.X(对)=1/2a<=3 -> a>=1/6
此时,f(x)在[3,+oo)单调递增 -> f(x)min=f(3)=10a-1
Ⅱ.X(对)=1/2a>3 -> a<1/6
f(x)min=f(1/2a)=a-1/4a
②.a>3时.
Ⅰ.当3<=x<a时,f(x)=ax^2-(a-x)=ax^2+x-a
X(对称轴)=-1/2a<0
故f(x)在[3,a)单调递增,f(x)min=f(3)=8a+3
Ⅱ.当x>=a时,f(x)=ax^2-(x-a)=ax^2-x+a
X(对称轴)=1/2a<a
所以f(x)在[a,+oo)单调递增
f(x)min=f(a)=a^3>8a+3
综上所诉,
当0<a<1/6时,f(x)的值域为[a-1/4a,+oo)
当1/6<=a<3时,f(x)的值域为[10a-1,+oo)
当a>3时,f(x)的值域为[8a+3,+oo)