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

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Answer 1 · 2017-01-08T03:06:41

#g(x)# has a minimum for #x=0#

#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#