Proving that $x^3 +1=15x$ has at most three solutions. in the interval [-4,4]. I need someone to check my work. Thanks! This is a 2 mark homework question by the way. I am not sure why am I using such a long way to prove it. Is there a way to shorten it or is there a shorter, more intuitive method?
Proving that $x^3 +1=15x$ has at most three solutions. in the interval [-4,4].

Let $f(x)=x^3+1-15x$

Suppose for a contradiction, that this equation has at least 4 solutions, $a,b,c,d$, such that $f(a)=0,f(b)=0,f(c)=0,f(d)=0$. Since f is continuous and differentiable on $x\in\mathbb{R}$, by the Rolle's Theorem, there exist a $c_1 \in (a,b) , c_2 \in (b,c),c_3 \in (c,d) $, such that $f^\prime(c_1)=0,f^\prime(c_2)=0,f^\prime(c_3)=0 $

$f^\prime(x)=3x^2-15$

Moreover, if $f^\prime(x)$ has 3 solutions, by the Rolle's Theorem, Since f is continuous and differentiable on $x\in\mathbb{R}$, there exist a $d_1 \in (c_1,c_2) , d_2 \in (c_2,c_3)$, such that $f^{\prime\prime}(d_1)=0,f^{\prime\prime}(d_2)=0$

$f^{\prime\prime}(x)=6x$

Moreover, if $f^{\prime\prime}(x)$ has 2 solutions, by the Rolle's Theorem, Since f is continuous and differentiable on $x\in\mathbb{R}$, there exist a $e_1 \in (d_1,d_2)$, such that $f^{\prime\prime\prime}(e_1)=0$

$f^{\prime\prime\prime}(x)=6$

This implies that $f^{\prime\prime\prime}(e_1)=0=6$ Hence, we have a contradiction. Without loss of generality, we can apply the steps to cases where $f(x)=x^3+1-15x$ has 5 or more solutions and still achieve a contradiction.  Therefore, the negation must be true, i.e $f(x)=x^3+1-15x$ has at most 3 solutions. 
 A: You only need to apply Rolle's Theorem once. A quadratic has at most two real solutions, this fact follows directly from the quadratic formula and doesn't need the Fundamental Theorem of Algebra. If you're willing to use the Fundamental Theorem of Algebra, then apply it directly to the cubic.
In response to OP's confusion.
The Fundamental Theorem of Algebra is a statement that a non-constant polynomial of degree $n$ has precisely $n$ roots (counting multiplicity) over the complex numbers. If we apply this theorem, then a cubic formula can have at most $3$ roots and we are in a direct contradiction if we have $4$. But as stated, the Fundamental Theorem of Algebra is overkill.
If you recall the quadratic formula, you will recall that if $x\in\mathbb{R}$ satisfies
$$ax^2 + bx + c = 0$$
then $x$ must also satisfy
$$x=\frac{-b\pm\sqrt{b^2 - 4ac}}{2a}$$
therefore we have (at most) only two possibilities for $x$
$$x=\frac{-b + \sqrt{b^2 - 4ac}}{2a}\ \ \ \text{or}\ \ \ x=\frac{-b-\sqrt{b^2 - 4ac}}{2a}$$
Therefore the quadratic formula is a direct proof that a degree two polynomial has at most $2$ distinct roots over the reals.
A: Degree 3 polynomials have exactly 3 roots, some of which could be complex. If it had more than 3 solutions in your interval, you get a contradiction with the fundamental theorem of algebra. 
A: As others have pointed out, the degree of the polynomial implies that any interval of $\mathbb{R}$ will contain at most three solutions for a cubic - by the Fundamental theorem of Algebra (which roughly says that a polynomial of degree $n$ has $n$ complex solutions, and thus at most $n$ real solutions).
If you were trying to show that $x^3-15x+1=0$ has exactly three real solutions in the interval, you could differentiate as you have done, show $f'=0$ has two solutions, that they lie in $[-4,4]$, and that $f(4)>0$, $f(-4)<0$ and (for instance) $f(1)<0$ and $f(-1)>0$, which implies $f(x)=0$ has three solutions in the interval.
If you really did mean at most, then forgetting the FTA, I suppose you could simply show $f'$ is $0$ exactly twice in the given interval, and hence $f$ can be $0$ at most thrice.
A: To be clear: Your solution seems to be correct.
However, as others have pointed out. You might not have to go that far. You for example find three solutions to the equation $3x^2 = 15 \Rightarrow x^2 = 5$. But you probably already know that there are only two solutions to this equation and so you have your desired contradiction.
A: I will give one answer, too. Perhaps this is enough intuitive.
Let $f(x):=x^3-15x+1$. We have to show, that $f$ has at most $3$ roots in the interval $[-4,4]$. Let $x\in[-4,4]$ with $f(x)=0$. Then $x\neq0$ and $x$ satisfies the equation
$$x(x^2-15)=-1.$$
If $x<0$ then $x^2-15>0$ and hence $x\in[-4,-\sqrt{15})$.
If $x>0$ then $x^2-15<0$ and hence $x\in(0,\sqrt{15})$.
So you know, if $f$ has roots in $[-4,4]$ then they must be in $[-4,-\sqrt{15})\cup(0,\sqrt{15}).$
Now $f'(x)=3x^2-15$ has exactly the roots $-\sqrt{5}, \,\sqrt{5}.$ Hence:
$f'$ has no root in $[-4,-\sqrt{15})$ $\Rightarrow$ $f$ has at most one root in $[-4,-\sqrt{15})$.
$f'$ has one root in $(0,\sqrt{15})$ $\Rightarrow$ $f$ has at most two root in $(0,\sqrt{15})$.
From these information you can see that $f$ has at most three roots in $[-4,4]$.
A: Rather than apply a cubic formula as some have pointed out, you can learn a method for finding solutions to cubic equations. It's easier to remember a method than a formula, where the slightest memory error can be a big deal.
Here is how the method goes, applied to this example. First, put all terms on one side.
$$x^3-15x+1=0$$
It is good that there is no quadratic term. Normally the first step is to make a change of variables to make the quadratic term go away, but here we don't need that step.
Next we introduce two variables to take the place of $x$. Yes, we are temporarily complicating things for a greater purpose. We bring in $u$ and $v$ with the condition that $x=u+v$.
$$(u+v)^3-15(u+v)+1=0$$
and multiply out the cube
$$u^3+3u^2v+3uv^2+v^3-15(u+v)+1=0$$
and group the middle two terms from the expanded cube
$$u^3+3uv(u+v)+v^3-15(u+v)+1=0$$
and bring the '$u+v$' terms together
$$u^3+(3uv-15)(u+v)+v^3+1=0$$
It would be great if $3uv-15$ equaled $0$, because then the middle section would be gone. Since we have two new variables to play with, we actually do have the freedom to choose $u$ and $v$ such that $3uv-15=0$, while still having a solution $x$ equal $u+v$.
$$u^3+v^3+1=0$$
Now that we have gone this route,$$3uv-15=0\implies v=5/u$$
and therefore 
$$u^3+(5/u)^3+1=0$$
after multiplying by $u^3$ and rearranging
$$u^6+u^3+125=0$$
which is just a surprise quadratic
$$(u^3)^2+u^3+125=0$$
And the quadratic formula gives 
$$u^3=\frac{-1+\sqrt{1-4(125)}}{2}=\frac{-1+i\sqrt{499}}{2}$$
(The other quadratic root must be $v^3$, since $u$ and $v$ play symmetric roles. So there's no need to keep track of the other root.)
Well, there are three complex cube roots to any nonzero complex number (oops! OP, if you don't get to use this fact, then this whole method has to stop here.)
So either $u=\alpha_1,\alpha_2$, or $\alpha_3$, the three cube roots of $\frac{-1+i\sqrt{499}}{2}$. In each case, $v=5/u$, and $x=u+v$. So there are at most three possibilities. Theoretically, some of them might actually be representing the same number, so without further calculation details, we would have to just say that there are at most three solutions.
