Answer
The equation $~~x^3-15x+c = 0~~$ has at most one root in the interval $[-2,2]$
Work Step by Step
$x^3-15x+c = 0$
Let $f(x) = x^3-15x+c$
$f'(x) = 3x^2-15$
Then $f'(x) \lt 0$ for all $x$ in the interval $[-2,2]$
The function $f(x)$ is continuous and differentiable for all $x$.
Let's assume that the equation has at least two roots $a$ and $b$ in the interval $[-2,2]$. Then $f(a) = f(b) = 0$. According to Rolle's Theorem, there is a number $k$ in the interval $(a,b)$ such that $f'(k) = 0$.
However, this contradicts the fact that $f'(x) \lt 0$ for all $x$ in the interval $[-2,2]$
Therefore, the equation has at most one root in the interval $[-2,2]$
