The minimum value of the function `|3-x|+9` is

Explanation

`y=x+1/x`
`dy/dx=d/dx(x)+d/dx(x^-1)`
`=1+(-x^-2)`
=1-1/x^2`
For y to be maximum or minimum,
`dy/dx=0`
or, `1-1/x^2=0`
Thus,` x=1 or -1`.
For xgt0 the only possible maximum or minimum is at x=1. At x=1,`y=x+1/x=2`
Near `x=0, y=x+1/x` is very large because of the term `1/x` For very large x agin y is very large because of the term x. Thus x=1 must correspond to a minimum. Thus, y has only a minimum for xgt0. This minimum occurs at x=1 and the minimum value of y is y=2.