Can this equation have 2 different solutions?

Question

Can this equation $x^3-12x=c$ have $2$ different solutions in $[-2,2]$? In $(-\infty,-2]$? In $[2,+\infty)$?

I said: Let the equation have 2 different solutions, one in $[-2,x_1]$ and one in $[x_1,2]$ and let $f(x)=x^3-12x-c,f(-2)<0$. According to Bolzano's theorem, $f(-2) \cdot f(x_1)<0 \implies f(x_1)>0 \implies f(2)<0 \implies -16-c<0$ and $32-c<0 \implies c>32$,but i that gives me nothing.

Answer 1

Picture a graph of $y = x^3 - 12x$; we need to know whether and where a horizontal line (at $y = c$) can intersect the graph twice. See if you can get a handle on what the graph looks like. Where is $y = x^3 - 12x$ increasing? Where is it decreasing?

Answer 2

The derivate of the function $$f(x)=x^3-12x-c$$ is $$f'(x)=3x^2-12=3(x-2)(x+2)$$

Hence the derivate is non-positive in the given interval because $x-2$ is non-positive and $x+2$ is non-negative. We have two isolated roots , namely $2$ and $-2$. Hence , $f(x)$ is strictly decreasing in the interval $[-2,2]$, hence it can have at most one root. Hence the equation can have at most one solution.

Answer 3

Let $f(x)=x^3-12x-c$. Then $f'(x)=3x^2-12$, which vanishes at $=2$ and $x=-2$. This means that $f$ is increasing at $(-\infty,-2]$ and $[2,\infty)$, and decreasing in $[-2,2]$.

There are five cases:

1. $f(-2)<0$. Then $f$ has a single root, and it is in $(2,\infty)$. This is for $c>16$.
2. $f(-2)=0$. Then $f$ has a double root at $x=-2$ and a single root in $(2,\infty)$. This is for $c=16$.
3. $f(-2)>0$ and $f(2)<0$. Then $f$ has three different roots, one in each interval of $(-\infty,-2)$, $(-2,2)$ and $(2,\infty)$. This is for $-16<c<16$.
4. $f(2)=0$. Then $f$ has a double root at $x=2$ and a single root at $(-\infty,-2)$. This is for $c=-16$.
5. $f(2)>0$. Then $f$ has a single root, and it is at $(-\infty,-2)$. This is for $c<-16$.

Remark: this answer is somewhat shortened. You may have to mention intermediate value theorem, explicitly compute the sign of $f'$, etc.

Answer 4

Assuming you want the solutions in $\mathbb{R}$ I believe you can answer this question using a divide a conquer approach, using different intervals and different values of c.

I also assume your question is in reality asking for the different values of which your equation will have solutions based on $c$. In any case this is irrelevant, you would need just to do the proper extrapolation of the results in negative case.

For a second, imagine the equation:

$x^3 - 12x = 0$

It has solutions for $x=0$ and $x=2\sqrt{3}$ and $x=-2\sqrt{3}$. So it have up to three solutions on $\mathbb{R}$ when $c=0$.

It is also convenient to have a visual: http://www.wolframalpha.com/input/?i=plot+x%5E3+-12x

It is clear that while $c$ "moves" up and down the former graph. The equation will have three solutions when $c$ does not shift up or down the graph beyond its maxima or minima.

Now we want to study the maxima and minima of that function. We can just use its derivate and equal to zero. Let $y=x^2 -12x -c$, then:

$y' = 3x^2-12$

Solving $y'=0$ gives $x=-2$ and $x=2$.

The maximum range of $c$ can be calculated as $y(2) = 8 - 24 = -16$, similar $y(-2) = 16$

if $c \in (-16,16)$ our eq. will only have 3 solutions. if $c=16$, or $c=-16$ our eq. will have 2 solutions. One that cuts and the other one which will be tangent to $y=0$ for the maxima or minima. Having just 1 solution for any other value of $c$.