Differentiation-please check my answers and help me on the last question
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Differentiation-please check my answers and help me on the last question

[From: ] [author: ] [Date: 11-11-15] [Hit: ]
f(1/3) = (1/27) + (1/9) - (1/3) - 1 = (-35/27) = -1.185.........
Show that the curve with equation y=x^3+x^2-x-1 has a stationary point when x=-1/3, and find the other x value where the curve has a stationary point.
dy/dx=3x^2+2x-1
so (3x-1)(x+1)
so x=1/3 and x=-1, correct?

ok just realised it's not but I don't understand why the sign are supposed to be the other way round, please explain,

Hence find the coordinates of the stationary points of the curve y=x^3+x^2-x-1

ok for this I have no idea where to start, please explain

thank you

-
3x^2+2x-1; a = 3, b = 2, c = 1

This is how the diagram is formed:
\ a*c /
[_] | [_]
/ b \

These two boxes on the sides are where your factors go...

\-3/
[3] | [-1] ===> (x+3)(x-1) ==(divide both by 3)==> (x+[3/3])(x-[1/3])
/2\

3 goes into 3 evenly, so that's that. 3 doesn't go into 1 evenly, so you move it in front of that X.

(x+1)(3x-1)
x=-1, 1/3

Now, take these two values and plug them into y=x^3+x^2-x-1 to find the stationary points of the curve.

f(-1) = 0
f(1/3) = (1/27) + (1/9) - (1/3) - 1 = (-35/27) = -1.185...
1
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