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Wednesday, 28 October 2015

STATIONARY POINTS EXAMPLE WITH AN INFLEXION AS WELL

QUESTION: Consider the curve y=x^3-3x^2-9x+27
(a) Find the coordinates of any stationary points and determine their nature.
(b) Find the coordinates of any points of inflexion (and verify using a y'' box that concavity changes)
(c) Get the x and y intercepts. (its possible here)
(d) Sketch it!
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Before we begin just find y' and y'' right away, factorise both of them ready for use later!

y'=3x^2-6x-9
  y'=3(x^2-2x-3)
  y'=3(x-3)(x+1)                                                (1)

y''=6x-6
y''=6(x-1)                                                          (2)


(a) For Stat Pts: solve y'=0.

 3(x-3)(x+1)=0                                      from (1)

\therefore x=-1, 3

For x=-1, y=(-1)^3-3(-1)^2-9(-1)+27=32

For x=3, y=(3)^3-3(3)^2-9(3)+27=0

Stationary Points are  (-1, 32)  and  (3,0).
Now determine their nature using a y' box for each point.

For (-1, 32).



 For (3, 0).



[REMARK: Instead of a y' box we could use y'' to test concavity at each x value, to determine the nature. This works only if y''\neq 0.
Since y''(-1)<0 it's concave down at x=-1, so it's a MAX.
Since y''(3)>0 it's concave up at x=3, so it's a MIN. ]

(b)  For possible inflexion, solve y''=0

6(x-1)=0
\therefore x=1.

When x=1y=(1)^3-3(1)^2-9(1)+27=16
Possible Inflexion is (1,16).

Verify that is an inflexion using y'' box.



(c)  For y intercept put x=0.  \therefore y=27 For x intercept put y=0.  \therefore 0=x^3-3x^2-9x+27. By inspection we see that x=-3,3.

[REMARK: its often the case that you can't easily find the x intercepts. Only do this as necessary or unless asked to do so in the question!]

(d) 

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