1. ## cubic equation

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'?

2. $\displaystyle 3x^2-x^3=k\Leftrightarrow x^3-3x^2+k=0$

Let $\displaystyle f(x)=x^3-3x^2+k, \ f:\mathbb{R}\to\mathbb{R}$

$\displaystyle f'(x)=3x^2-6x$

$\displaystyle f'(x)=0\Rightarrow x_1=0, \ x_2=2$

$\displaystyle f(0)=k, \ f(2)=k-4, \ \lim_{x\to-\infty}f(x)=-\infty, \ \lim_{x\to\infty}f(x)=\infty$

The Rolle's sequence is: $\displaystyle -\infty \ \ k \ \ k-4 \ \ \infty$

To have 3 real solutions the following inequalities must hold:

$\displaystyle \left\{\begin{array}{ll}k\geq 0\\k-4\leq 0\end{array}\right.\Rightarrow k\in[0,4]$

3. Draw the graph of the left hand side of the equation. Where you can draw a horizontal line such that the line intersects the graph three times is a value of k that you want - find all such values.

(drawing is meant for illustration only - it will need to be worked out mathematically. hint: consider turning points)

4. Originally Posted by red_dog
$\displaystyle 3x^2-x^3=k\Leftrightarrow x^3-3x^2+k=0$

Let $\displaystyle f(x)=x^3-3x^2+k, \ f:\mathbb{R}\to\mathbb{R}$

$\displaystyle f'(x)=3x^2-6x$

$\displaystyle f'(x)=0\Rightarrow x_1=0, \ x_2=2$

$\displaystyle f(0)=k, \ f(2)=k-4, \ \lim_{x\to-\infty}f(x)=-\infty, \ \lim_{x\to\infty}f(x)=\infty$

The Rolle's sequence is: $\displaystyle -\infty \ \ k \ \ k-4 \ \ \infty$

To have 3 real solutions the following inequalities must hold:

$\displaystyle \left\{\begin{array}{ll}k\geq 0\\k-4\leq 0\end{array}\right.\Rightarrow k\in[0,4]$
or red_dog will just give you the answer...

5. thank you both