Please help curve/differentiation Q?

Question

The Q says
The Curve C has equation y = x^3 - 3x^2
(A)Find the co-ordinates of the stationary points of C and determine the nature of these points

(B) Sketch C (obviously cannot do)

C) Find the range of values of k for which there are three real and distinct solutions of the equation x^3 - 3x^2 = k

Mr Maths, thanks for your help but you've read the question wrong, the + is a - on the equation

Answer 1

Here is how to do it:

1. You differentiate the curve : y=x^3+3x^2. dy/dx=3x^2+6x

2. Then you must put the equation = 0. Therefore, it is dy/dx=0.
Hence, 0=3x^2+6x.

3. Factorise the equation: x(3x+6)=0 or 3x(x+2)=0.
Therefore 3x=0 or x+2=0, therefore x=-2 and x=0.

4. Put these values into the original equation if x=0, y=0 (you do the arithmetic!) and if x=-2,y=4. Stat. pts are (0,0) and (-2,4)

(B) Sketch the curve. Advice: If you do not have a clue about the shape of the curve, then plot the curve using a few points x=-2 to x=2. It might be a good idea to find out the nature of the stat. points d2y/dx2 (differentiate again)<0 max. d2y/dx2>0 min, d2y/dx2=0 (can be a point of inflexion but do not assume this).

(C) Need to get back to the A level maths books for that one!

Answer 2

A) Differentiate
y' = 3x^2 - 6x
Stationary pts when y' = 0
therefore 3x(x - 2) = 0 which implies x = 0, x = 2
When x = 0 y = 0 (max), when x = 2 y = -4 (min)
y'' = 6x - 6
if x = 0 y'' < 0 max
if x = 2 y'' >0 min

C) 0

Answer 3

a) f(x) = x³ - 3x²
f `(x) = 3x² - 6x =3x(x - 2) =0 for stationary values
Thus x = 0,x =2 for stationary values
f(0) = 0 and f(2) = -4
Turning points are (0,0) and (2,-4)

b) f ``(x) = 6x - 6
f "(0) = -6 a negative value thus (0.0) is a maximum turning point
f " (2) = 12 - 6 = 6 a positive value thus (2,-4) is a minimum turning point

Could then sketch curve which passes thro,(0,0),(2,-4) and (3,0)

Answer 4

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Answer 5

y' = dy/dx = 3x^2 - 6x
solve y' = 0 for stat points