Question

求函数y=2x²-ax+1,x∈ [0,1]的最小值g(a)和最大值h(a)

Answer 1

抛物线y=f(x)=2x²-ax+1开口向上，对称轴为直线x=a/4.
a/4∈[0,1]时，g(a)=f(a/4)=1-a^2/8;
a/4<0时，g(a)=f(0)=1;
a/4>1时，g(a)=f(1)=3-a.
a/4<=1/2时，h(a)=f(1)=3-a;
a/4>1/2时h(a)=f(0)=1.
综上，g(a)={1,a<0;
{1-a^2/8,0<=a<=4;
{3-a,a>4.
h(a)={3-a,a<=2;
{1,a>2.