What are the critical points of #f(x) = x^2 + 2/x#?

Question

6338 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2016-12-02T14:12:13

#(1, 3)# is the only critical point.

The critical points will occur when the derivative equals #0# or is undefined .

#f'(x) = 2x + (0 xx x - 1 xx 2)/(x)^2#

#f'(x) = 2x - 2/x^2#

The derivative will be undefined at #x = 0#. However, the function is also undefined at #x = 0#, so this is not a critical point.

#0 = 2x - 2/x^2#

#0 = (2x^3 - 2)/x^2#

#0 = 2x^3 - 2#

#2 = 2x^3#

#1= x^3#

#x = 1#

The corresponding y-coordinate is

#f(1) = 1^2 + 2/1 = 3#

Hence, the only critical point is at #(1, 3)#.

Hopefully this helps!