## Calculus: Early Transcendentals 8th Edition

The equation $~~x^3-15x+c = 0~~$ has at most one root in the interval $[-2,2]$
$x^3-15x+c = 0$ 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 $[-2,2]$