The function `f(x)=x/2+2/x` has a local minimum at `x=2` (b) `x=-2` `x=0` (d) `x=1`

Options

A. x = - 2
B. x = 0
C. x = 1
D. x = 2

Answer

D

Explanation

Given, `f(x)=(x)/(2)+(2)/(x)`
`rArr f'(x)=(1)/(2)-(2)/(x^(2))`
Put f'(x) = 0 for maxima or minima.
`rArr (1)/(2)-(2)/(x^(2))=0`
`rArr x^(2)=4`
`rArr x=pm 2`
Now, `f''(x)=(4)/(x^(3))`
`f''(2)=(4)/(8)=(1)/(2) gt 0`
`and f''(2)=-(4)/(8)= -(1)/(2) lt 0`
`therefore` f(x) is minimum at x = 2.