pc4pc said:
..finding the S.Ps is what i'm having trouble with for these Q's!! =/
General rules:
Stationary points exist at the co-ordinates for which
f'(x) = 0 (here the tangent is horizontal).
Let's assume a stationary point occurs at x=a. Finding f(a) will give you the corresponding y co-ordinate. Often you will need to determine the nature of these points to assist the sketch. There are two ways of doing this:
1. Find f"(a). If
f"(a) > 0, the point is
minimum turning point; if
f"(a) < 0, the point is a
maximum turning point.
2. Use f'(x) to test one value from either side of x=a. Draw up a table with the three values and you should see a trend in their respective f'(x) values of either
+ve 0 -ve (increasing to flat to decreasing, meaning a
maximum) or
-ve 0 +ve (decreasing to flat to increasing, meaning a
minimum).
[Note: +ve 0 +ve and -ve 0 -ve trends signify horizontal points of infelxion.] Use this method if the expression for f"(x) is too labourious to find.
Points of inflexion exist at co-ordinates for which
f"(x) = 0 (this signifies a change in concavity in most circumstances, but not always). After finding a point of inflexion, you
MUST test the f"(x) of one value from either side of the point to confirm the change in concavity (it will also give you a idea of the shape of the curve). A lot of people tend to forget about this and you will lose marks if you don't remember. The reason for doing this is because there are some circumstances where f"(x) = 0 can occur for stationary turning points, e.g. y = x^n where integer n > 3.
Horizontal points of inflexion are both stationary points & points of inflexion, so at these points
f'(x) = 0 AND
f"(x) = 0.
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For example:
1. f(x) = 3x^4 + 4x^3 - 12x^2 - 1,
showing all stationary points.
f'(x) = 12x^3 + 12x^2 - 24x
[factorising this fully with make solving f'(x) =0 easier]
= 12 * (x^3 + x^2 - 2x)
= 12x * (x^2 + x - 2)
= 12x * (x - 1) (x + 2)
Stat. pts. occur at f'(x) = 0
i.e. 12x * (x - 1) (x + 2) = 0
so, x = -2, 0, 1
f(-2) = -33
f(0) = -1
f(1) = -6
Therefore, the stat. pts. are (-2, -33) (0, -1) and (1, -6).
From f'(x) = 12x^3 + 12x^2 - 24x
f"(x) = 36x^2 + 24x -24
f"(-2) = 72 > 0 so (-2, -33) is a min pt.
f"(0) = -24 < 0 so (0, -1) is a max pt.
f"(1) = 36 > 0 so (1, -6) is min pt.
You should also find and label the x- and y- intercepts on your sketch.
2. find all stationary points and inflexions on the curve
y = (2x + 1)(x - 2)^4.
y' = [2 * (x - 2)^4] + [(2x + 1) * 4 * (x - 2)^3]
= [2 * (x - 2)^3] [(x - 2) + 2 * (2x + 1)]
= [2 * (x - 2)^3] [x - 2 + 4x + 2]
= 10x * (x - 2)^3
Stat. pts. occur at y' = 0
i.e. 10x * (x - 2)^3 = 0
so x = 0 --> y = 16
& x = 2 --> y = 0
Therefore, the stat. pts. are (0, 16) and (2, 0).
From y' =10x * (x - 2)^3
y" = [10 * (x - 2)^3] + [10x * 3 * (x - 2)^2]
= [10 * (x - 2)^2] [(x - 2) + 3x]
= 10 * (4x - 2) * (x - 2)^2
= 20 * (2x - 1) * (x - 2)^2
y"(0) = -80 < 0 so (0, 16) is max pt.
y"(2) = 0
[You MUST test the y" of points on either side of x = 2 and in doing that you will realise there is no change in concavity and (2, 0) is a min pt.]
Pts. of infl. occur at y" = 0
i.e. 20 * (2x - 1) * (x - 2)^2 = 0
so x = 1/2 --> y = 81/8
[Again, you MUST test the y" of points on either side of x = 1/2 to confirm]
x = 2 is ignored because (2, 0) was shown to be a min pt.
Therefore, the pt. of infl. is (1/2, 81/8)
Coincidentally, the stat. pts. are also the x- and y- intercepts.
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I hope this has clarified the problem for you. You should be able to use the same methods for most 2U curve sketching questions under applications of calculus. Good luck.
