Explain why the equation
$$x^3 - 15x +1 = 0$$
has at least three solutions in the interval [-4,4].
My thoughts:
$$f(x) = x^3 - 15x + 1$$ $$f(-4) = -3 $$ $$f(4) = 5 $$ $$f(-4) < 0 < f(4)$$ Therefore, by IVT, there exist some $c \in [-4,4]$ such that $f(c)=0$ exist.
However, I am unable how to prove there are at least three solutions...
On
This proves it has at least one solution.
Hint:
Show $f(x)$ has a local maximum and a local minimum on $[-4,4]$, and that the maximum is positive and the minimum is negative.
On
$$f(-4)=-3$$ $$f(0)=1$$ $$f(1)=-13$$ $$f(4)=5$$
By IVT, there are at least three roots
one in $(-4,0)$ ,
one in $(0,1)$,
and one in $(1,4)$.
A polynom with degee $3$ cannot have more than three roots.
If it had four roots, the derivative will have three roots (By Rolle).
but $$f'(x)=3x^2-15$$
which have only two roots. So, $$x^3-15x+1=0$$ has exactly three real solutions.
Another approach is to identify the turning points of $f(x)$ by setting $f'(x)=3x^2-15=0$ so that $x=\pm \sqrt 5$ these lie between $x=\pm 4$ and the general shape of the curve gives a local maximum at $x=-\sqrt 5$. Since $f(0)=1$ it is easy to see that the value at the maximum is positive. Continue analysis to identify what you need.