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red and white kop!

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the equation 3x^2 - x^3 = k has three real solutions for k
write down the set of possible values for k

i dont understand: can't k = anything as long as x remains real? and what is meant by '3x^2 - x^3 = k has three real solutions for k'?
 
red and white kop! said:
the equation 3x^2 - x^3 = k has three real solutions for k
write down the set of possible values for k

i dont understand: can't k = anything as long as x remains real? and what is meant by '3x^2 - x^3 = k has three real solutions for k'?
Click to expand...

There could be analytical way to solve this problem - but I do not know that.

My way of solving the problem is:

sketch y = x^3 - 3x^2 + k with k = 0, ±1, ±2,.... the answer jumps out.
 
Since there are three x-intercepts the equation has three real solutions.
 
red and white kop! said:
um no i dont think thats the answer im looking for, must be sumthing with the roots no?
Click to expand...
Did you sketch the curves as I suggested?

If you did - what did you observe as you change the values of 'k' - from positive to negative?
 
If 0<k<4, you'll get 3 solutions

If k =0 or k =4, you'll get 2 solutions.

If k < 0 or if k >4, you'll get one solution.

Note: I did this by trial and error, is there a formula for doing this?
 
BigGlenntheHeavy said:
If 0<k<4, you'll get 3 solutions

If k =0 or k =4, you'll get 2 solutions.

If k < 0 or if k >4, you'll get one solution.

Note: I did this by trial and error, is there a formula for doing this?
Click to expand...

I did the same graphically - I admitted before that I do not know the analytical solution.
 
I think I figured it out.

Let f(x) = x^3-3x^2+k, k a constant. Hence we have a cubic family of functions

f ' (x) = 3x^2-6x = 3x(x-2), x=0 (rel max), x= 2 (rel min), point of inflectiion immaterial.

Ergo f(0) = k and f(2) = k-4. now in order to have three solutions the rel max must be above the x axis and the rel min must be below the x axis. hence k > 0 and k-4 < 0. Therefore 0<k<4. QED.
 
Right on, Subhotosh Khan, if you think about it, you have a family of cubics in which you are only transposing the function up or down on the y axis, hence one wants to find that "window" where the function intersects the x axis three times. Ergo, for this to happen the rel max must be above the x axis and the rel min must be below the x axis.
 
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