What are the local extrema of #f(x)= x^3-x+3/x#?

Question

1669 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2016-12-06T17:16:15

#x_1= -1# is a maximum
#x_2= 1# is a minimum

First find the critical points by equating the first derivative to zero:

#f'(x) = 3x^2-1-3/x^2#

#3x^2-1-3/x^2 = 0#

As #x!=0# we can multiply by #x^2#

#3x^4-x^2-3=0#

#x^2=frac(1+-sqrt(1+24)) 6 #

so #x^2=1# as the other root is negative, and #x=+-1#

Then we look at the sign of the second derivative:

#f''(x) = 6x+6/x^3#

#f''(-1) = -12 <0#

#f''(1) = 12>0#

so that:

#x_1= -1# is a maximum
#x_2= 1# is a minimum

graph{x^3-x+3/x [-20, 20, -10, 10]}