What are the local extrema of #f(x)= (x^5-x^2-4)/(x^3-3x+4)#?

Question

1212 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2016-02-10T16:50:06

Local maximum #~~ -0.794# (at #x~~ -0.563#) and local minima are #~~ 18.185# (at #x~~ -3.107#) and #~~ -2.081# (at #x~~0.887#)

#f'(x) = (2x^7-12x^5+21x^4+15x^2-8x-12)/(x^3-3x+4)^2#

Critical numbers are solutions to

#2x^7-12x^5+21x^4+15x^2-8x-12 = 0#.

I do not have exact solutions, but using numerical methods will find real solutions are approximately:

#-3.107#, #- 0.563# and #0.887#

#f''(x) = (2x^9-18x^7+14x^6+108x^5-426x^4+376x^3+72x^2+96x-104)/(x^3-3x+4)^3#

Apply the second derivative test:

#f''(-3.107) > 0#, so #f(-3.107) ~~ 18.185# is a local minimum

#f''(- 0.563) < 0#, so #f(- 0.563) ~~ -0.794# is a local maximum

#f''(0.887) > 0#, so #f(0.887) ~~ -2.081# is a local minimum