The question is as follows:
Show that the equation $x^3 -15x +c =0$ has at most one root in the
interval $ [-2, 2]$
I had an idea on how to solve this problem, although it is completely different to the solution from the textbook. I hoped someone might be able to validate my working or to show me the error of my ways.
I reason that if the maxima and maxima x values of a function correspond to the roots on the graph of a derivative, then maybe the roots of a function correspond to the maximas and minimas of the graph of the derivative?
Because f(x) is continuous and differentiable on the interval, we can apply the Mean Value Theorem to the derivative of the function:
$$f''(c) =\frac{f(b)-f(a)}{b-a}\ $$
$$ 6(c) = \frac{-3+3}{0}\ $$
$$ 6c = 1 $$
$$ c = \frac{1}{6}\ $$
Because there is only one critical value, c, it can be said that f(x) has at most one root.
Any feedback would be much appreciated. Thanks in advance
Best Answer
Assume there are $2$ or more real roots, then by the Rolle's theorem: $f'(c) = 0$ in $[-2,2]$, but $f'(c) = 3c^2 - 15 = 3(c^2-5) < 0$ since $|c| \leq 2$. Thus it can't happen.