Question

Determine the local max and/or min and intervals of increase and decrease for the function f(x)=√(x^2 - 2x +2) ?

Answer 1

`f` is decreasing in `(-oo,1]` and increasing in `[1,+oo)` so `f` has a local and global `min` at `x_0=1` , `f(1)=1`
`-> f(x)>=f(1)=1>0` , `x` `in` `RR`

Explanation:

`f(x)=sqrt(x^2-2x+2)` , `D_f=RR`

`AA` `x` `in` `RR`,

`f'(x)=((x^2-2x+2)')/(2sqrt(x^2-2x+2)` `=`

`(2x-2)/(2sqrt(x^2-2x+2)` `=`

`(x-1)/(sqrt(x^2-2x+2)`

with `f'(x)=0 <=> (x=1)`

• `x` `in` `(-oo,1)`, `f'(x)<0` so `f` is decreasing in `(-oo,1]`
• `x` `in` `(1,+oo)` , `f'(x)>0` so `f` is increasing in `[1,+oo)`

`f` is decreasing in `(-oo,1]` and increasing in `[1,+oo)` so `f` has a local and global `min` at `x_0=1` , `f(1)=1`
`-> f(x)>=f(1)=1>0` , `x` `in` `RR`

Graphical help
graph{sqrt(x^2-2x+2) [-10, 10, -5, 5]}