Let `f(x)=x^2+(1/x^2)` and `g(x)=x-1/x` `x in R-{-1,0,1}`. If `h(x)=(f(x)/g(x))` then the local minimum value of `h(x)` is: (1) 3 (2) `-3` (3) `-2sqrt(2)` (4) `2sqrt(2)`

Options

A. `2sqrt(2)`
B. 3
C. -3
D. `-2sqrt(2)`

Answer

1

Explanation

`h(x) =(x^(2)+(1)/(x^(2))/(x-(1)/(x))=(x-(1)/(x))^(2)+2)/(x-(1)/(x))=(x-(1)/(x))+(2)/(x-(1)/(x))`
Let `x-(1)/(x)=z`
For local minimum value of `h(x) let zgt0`
`therefore h(x) =z+(2)/(z)=sqrt(z)-sqrt(2)/(z)^(2)+2sqrt(2)ge2sqrt(2)`
So local minimum value of h(x) is `2sqrt(2)`
if `zlt0` then
`h(x)=-[-a-(2)/(z)]=-[sqrt(-z)-sqrt(2)/(-z)]^(2)-2sqrt(2)le-2sqrt(2)`
So `-2sqrt(2)` is locall maximum value of h(x)