Question

How do you find the extrema for `g(x) = sqrt(x^2 + 2x + 5)`?

Answer 1

`g(x)` has no maximum and a global and local minimum in `x=-1`

Explanation:

Note that:

`(1) " "x^2+2x+5 = x^2+2x+1+4 = (x+1)^2+4 > 0`

So the function

`g(x) = sqrt(x^2+2x+5)`

is defined for every `x in RR`.

Besides as `f(y) = sqrty` is a monotone increasing function, then any extremum for `g(x)` is also an extremum for:

`f(x) = x^2+2x+5`

But this is a second order polynomial with leading positive coefficient, hence it has no maximum and a single local minimum.

From `(1)` we can easily see that as:

`(x+1)^2 >= 0`

and:

`x+1=0`

only when `x=-1`, then:

`f(x) >= 4`

and

`f(x) = 4`

only for `x=-1`.

Consequently:

`g(x) >= 2`

and:

`g(x) = 2`

only for `x=-1`.

We can conclude that `g(x)` has no maximum and a global and local minimum in `x=-1`

Answer 2

`g(x)=sqrt(x^2+2x+5)` , `x` `in` `RR`

We need `x^2+2x+5>=0`

`Δ=2^2-4*1*5=-16<0`

`D_g=RR`

`AA` `x` `in` `RR`:

`g'(x)=((x^2+2x+5)')/(2sqrt(x^2+2x+5))` `=`

`(2x+2)/(2sqrt(x^2+2x+5))` `=`

`(x+1)/(sqrt(x^2+2x+5)>0)`

`g'(x)=0` `<=>` `(x=-1)`

• For `x<-1` we have `g'(x)<0` so `g` is strictly decreasing in `(-oo,-1]`

• For `x>` `-1` we have `g'(x)>0` so `g` is strictly increasing in `[-1,+oo)`

Hence `g(x)>=g(-1)=2>0` , `AA` `x` `in` `RR`

As a result `g` has a global minimum at `x_0=-1` , `g(-1)=2`