## Calculus 10th Edition

Rolle's Theorem can be applied; $c=\sqrt{5}-2.$
While $f(x)$ is not continuous at $x=-2$, the specified interval does not include $x=-2$ and hence $f(x)$ is continuous for all the values of $x$ in the closed interval and differentiable at every value of $x$ in the open interval. $f(-1)=f(3)=0.$ Since $f(x)$ is continuous over $[-1 ,3 ]$ and differentiable over $(-1, 3)$, applying Rolle's Theorem over the interval $[-1, 3]$ guarantees the existence of at least one value $c$ such that $-1\lt c\lt3$ and $f'(c)=0.$ Using the quotient rule: $f’(x)=(\frac{u(x)}{v(x)})'=\frac{u'(x)v(x)-v'(x)u(x)}{(v(x))^2}$ $u(x)=x^2-2x-3; u'(x)=2x-2$ $v(x)=x+2; v'(x)=1$ $f'(x)=\dfrac{(2x-2)(x+2)-(x^2-2x-3)(1)}{(x+2)^2}=\dfrac{x^2+4x-1}{(x+2)^2}$ $f'(x)=0\to \dfrac{x^2+4x-1}{(x+2)^2}=0\to x^2+4x-1=0\to$ Using the Quadratic equation, we get $c=\sqrt{5}-2$ or $c=-\sqrt{5}-2$ (which is rejected since $c\gt-1$)