reposting this question because no one could do it!!! MATHS!?

Question

The curve c has equation: y=x^3 -3x^2

a.) Find Co-ordinates of stat points of C and determine nature.
b.) Sketch C
c.) Find the range of values of k for which there are 3 real and distinct solutions of the equation x^3 - 3x^2 = k

Any help with any of the questions would be great?

Answer 1

a/

y ' =3x^2 - 6x = 0--->x=0 or 2

maximum (0,0)

min (2,-4)

b/

now sketch

c/

-4

Answer 2

I think that sketching this is the first thing you want to do.



The first thing I would do is find the zeros, that is, find where y = 0.

0 = x^3 - 3x^2

0 = x^2(x - 3)

so where the 'y' value of the graph is zero where x = 0 or 3.



Next I would plot a few points to see where they land. On this one, you would want to use 0 and 3 as a base. I would plot all the plots between -5 and 5. This should tell you the nature, or the end behavior and shape of this curve.



To find the range of this function you would look at how this thing behaves at each x value. This one is continuous, meaning that at each and every value of x, a y value exists. If there were a square root, or an x in a denominator position within the function, then there will be discontinuities (that is, at certain values of x the values of y do not exist.)



There are ways to do this easier using calculus, but I don't know what level of maths you are at.



Hope this helps.

Answer 3

a) Assuming stat points means critical points

dy/dx = 3x^2 - 6x

0 = 3x^2 - 6x

0 = 3x(x - 2)

3x = 0 or x - 2 = 0

x = 0 or x = 2

d2y/dx2 = 6x - 6

d2y/dx2(0) = - 6 < 0 so x = 0 is a local maximum

d2y/dx2(2) = 12 - 6 = 6 > 0 so x = 2 is a local minimum

y(0) = 0

(0,0) is a local maximum

y(2) = 8 - 12 = -4

(2,-4) is a local minimum



b) Use above information to graph



c) There will be 3 real solutions where a horizontal line intersects y = x^3 - 3x^2 three times.

y(0) = 0. We need the other point such that y = 0

0 = x^3 - 3x^2

0 = x^2(x - 3)

The other point is x = 3

y(2) = -4. We need the other point such that y = -4

x^3 - 3x^2 = 4

The other point is x = -1

The range of values of k for which there are 3 real, distinct solutions is [-1,0) U (0,2) U (2,3].

Note that there are only two solutions at x = 0 and x = 2 so they are excluded.

Answer 4

Stat points are when the derivative is equal to zero.

dy/dx = 3x² - 6x = 0

So there are stationary points when 3x² = 6x, or x² = 2x. That is, when x = 0 or x = 2.



To determine the nature, find the second derivative.d²y/dx² = 6x - 6. Now put the values of x back in. If d²y/dx² is positive, it's a minimum point. If it's positive, it's a maximum point. If it's zero, it's a point of inflection. So for x = 0, d²y/dx² = -6, so it's a maximum point. For x = 2, d²y/dx² = 6, so it's a maximum.



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Sketching this jazzy function:

Plot the points y = x³ - 3x² for x = 0 and x = 2. So for these, y = 0 and y = -4 respectively. Remembering their nature from just before, draw a curve coming from minus infinity to the maximum point (0,0), then back down to the point (2,-4), then up to infinity from there.



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EDIT: The people above me are right here, look at the graph, x³ - 3x² = k is essentially asking 'for a line of y = k, which values can k take such that it intersects the curve you just drew three times?'.

Answer 5

a)



find the derivative function of this curve which is y'=3x^2-6x=3x(x-2)

y'>0 when x<0 and x>2

y'<0 ehrn 0
y'=0 when x=0 and x=2



so the local maximum is (0;0)

the local minimum is (2;-4)



b)



(C) will rise from -infinity to (0;0) then down to (2;-4) then to infinity



c)



x^3-3x^2=k has 3 distinct solution when -4<=k<=0



see it when you sketch the graph ( between the local maximum and the local minimum)