Question

求y=|x^2-3x+2|在[-3,4]上的最大值,最小值

Answer 1

y=|x^2-3x+2| = |(x-1)(x-2)| =

零点为x1=1,x2=2,图像顶点在x=(x1+x2)/2=3/2处

f(-3) = |(-3-1)(-3-2)| = 20

f(4) = |(4-1)(4-2)| = 6

f(3/2) = |(3/2-1)(3/2-2)| = 1/4

最大值f(-3)=20

最小值f(1)=f(2)=0