Thread: f(x) graph based on derivative graph help
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Old January 1st, 2009, 12:41 AM
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Default Maximum and minimum
Hello OnMyWayToBeAMathProffesor

First, note that your answers to a and b are contradictory: you can't have a point of inflexion that's also a maximum. So, let's review what you need in order to determine which is which:
  • At a maximum or minimum turning point, f'(x) =0
  • At a maximum, f'(x) goes from being positive on the left of the turning point to being negative on the right of it. So f'(x) is decreasing, and hence f''(x) is negative. And conversely...
  • At a minimum, f''(x) is positive.
  • At a point of inflexion f''(x)=0
So, from the table, the only point at which f'(x) =0 is where x=1, and at this point f''(x)=0. So:

a) There are no maximum or minimum points in this interval

b) There is a point of inflexion where x =1

c) As far as the sketch is concerned:
  • For -3<x<-1, the gradient of the graph is positive, and it's increasing (because both f' and f'' are positive), and tends to infinity as x tends to -1. So there's going to be an asymptote at x=-1.
  • For -1<x<1, gradient is negative but the graph is levelling out since f'' is positive and hence the gradient is increasing (i.e. becoming less negative)
  • At x=1, the graph is horizontal, but the gradient starts to go negative again after this point - hence a point of inflexion.
  • For x>1, the gradient continues to decrease, and the graph therefore becomes steeper.
You should be able to sketch the graph now. (The way I would do it on MHF is to create a .jpg either using a drawing package or scanning a hand-drawn graph, and attach it using the 'Manage Attachments' button.)

Hope that helps.

Grandad
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