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
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...
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...