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. 

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.
