Question

如果二次函数f(x)=3x的平方+mx+1在区间（负无穷,-1/3】上是增函数,则函数f(x)在区间【-1,1】上的最大

Answer 1

f(x)=3x^2+mx+1

f'(x) = 6x+m

f'(-1/3) = -2+m >0

m > 2

f'(x)=0

6x+m=0

x = -m/6

f''(x)=6 (min)

f(-1) = 3-m+1

f(1) = 3+m+1 > f(-1) ( m > 2 )

max f(x) = f(1) = 4+m #