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

1 Answer
Jan 8, 2017

g(x) has a minimum for x=0

Explanation:

g(x) = x^2+1

g'(x) = 2x

g''(x) = 2

We find the critical point as solution of the equation:

g'(x) = 0

2x = 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(x) is a second order polynomial so it has a single local extreme.
(ii) x^2>=0 so g(x) >=1 and g(0) = 1