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

May 4, 2018

$g \left(x\right)$ has no maximum and a global and local minimum in $x = - 1$

#### Explanation:

Note that:

$\left(1\right) \text{ } {x}^{2} + 2 x + 5 = {x}^{2} + 2 x + 1 + 4 = {\left(x + 1\right)}^{2} + 4 > 0$

So the function

$g \left(x\right) = \sqrt{{x}^{2} + 2 x + 5}$

is defined for every $x \in \mathbb{R}$.

Besides as $f \left(y\right) = \sqrt{y}$ is a monotone increasing function, then any extremum for $g \left(x\right)$ is also an extremum for:

$f \left(x\right) = {x}^{2} + 2 x + 5$

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

From $\left(1\right)$ we can easily see that as:

${\left(x + 1\right)}^{2} \ge 0$

and:

$x + 1 = 0$

only when $x = - 1$, then:

$f \left(x\right) \ge 4$

and

$f \left(x\right) = 4$

only for $x = - 1$.

Consequently:

$g \left(x\right) \ge 2$

and:

$g \left(x\right) = 2$

only for $x = - 1$.

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

May 4, 2018

$g \left(x\right) = \sqrt{{x}^{2} + 2 x + 5}$ , $x$$\in$$\mathbb{R}$

We need ${x}^{2} + 2 x + 5 \ge 0$

Δ=2^2-4*1*5=-16<0

${D}_{g} = \mathbb{R}$

$\forall$$x$$\in$$\mathbb{R}$:

$g ' \left(x\right) = \frac{\left({x}^{2} + 2 x + 5\right) '}{2 \sqrt{{x}^{2} + 2 x + 5}}$ $=$

$\frac{2 x + 2}{2 \sqrt{{x}^{2} + 2 x + 5}}$ $=$

$\frac{x + 1}{\sqrt{{x}^{2} + 2 x + 5} > 0}$

$g ' \left(x\right) = 0$ $\iff$ $\left(x = - 1\right)$

• For $x < - 1$ we have $g ' \left(x\right) < 0$ so $g$ is strictly decreasing in $\left(- \infty , - 1\right]$

• For $x >$$- 1$ we have $g ' \left(x\right) > 0$ so $g$ is strictly increasing in $\left[- 1 , + \infty\right)$

Hence $g \left(x\right) \ge g \left(- 1\right) = 2 > 0$ , $\forall$$x$$\in$$\mathbb{R}$

As a result $g$ has a global minimum at ${x}_{0} = - 1$ , $g \left(- 1\right) = 2$ 