求函数f(x)=p2x2(1-x)p(p∈N+),在[0,1]内的最大值.
f''(x)=p2x(1-x)p-1[2-(2+p)x], 令f''(x)=0,得x=0,x=1,x=2/(2+p), 在[0,1]上,f(0)=0,f(1)=0,f[2/(2+p)]=4[p(2+p)]p+2 所以[f(x)]max=4[p(2+p)]2+p
求函数f(x)=p2x2(1-x)p(p∈N+),在[0,1]内的最大值.
f''(x)=p2x(1-x)p-1[2-(2+p)x], 令f''(x)=0,得x=0,x=1,x=2/(2+p), 在[0,1]上,f(0)=0,f(1)=0,f[2/(2+p)]=4[p(2+p)]p+2 所以[f(x)]max=4[p(2+p)]2+p