# How do you find the local extremas for g(x) = x^2 + 1?

Jan 8, 2017

$g \left(x\right)$ has a minimum for $x = 0$

#### Explanation:

$g \left(x\right) = {x}^{2} + 1$

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

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

We find the critical point as solution of the equation:

$g ' \left(x\right) = 0$

$2 x = 0$

${x}_{c} = 0$

As the second derivative is constant and positive this critical point is a local minimum.

The same conclusion can be drawn by direct inspection as:

(i) $g \left(x\right)$ is a second order polynomial so it has a single local extreme.
(ii) ${x}^{2} \ge 0$ so $g \left(x\right) \ge 1$ and $g \left(0\right) = 1$