Question

已知函数f(x)=x^2+x/x-1其定义域[2,3],求函数f(x)的最小值

Answer 1

f(x)=x^2+x/x-1

=(x^2-x+2x)/(x-1)

=x+2+2/(x-1)

=(x-1)+2/(x-1) +3

≥2×√[(x-1)·2/(x-1)] +3

=3+2√2

.

f'(x)=[x+2+2/(x-1)]'

=1-2/(x-1)^2

则f'(1+√2)=0.

当x∈[2,1+√2]时,f'(x)