已知fx=kx³-3(k+1)x²-k²+1(k>0)若fx的单调递减区间为[0,4]求k

Question

已知f x=kx³-3(k+1)x²-k²+1(k>0)若f x的单调递减区间为[0,4]求k

Answer 1

f x=kx³-3(k+1)x²-k²+1
f'(x)=3kx²-6(k+1)x
=3x[kx-2(k+1)]
令f'(x)=0解得x=0,或x=2(k+1)/k
(k≠0)
∵f x的单调递减区间为[0,4]
∴x1=0,x2=4是f'(x)=0的根
那么2(k+1)/k=4
∴k=1
此时f'(x)=3x(x-4)f(x)单调递减区间为[0,4]
∴k=1