Range Rover!! :p

The minimum and maximum of a differentiable function occur either at the endpoints of its domain or where the derivative equals zero. The domain of $f(x)$ is $\{x|-1\leq x \leq 3\}$ . $f'(x)=\frac{1}{2\sqrt{x+1}}-\frac{1}{2\sqrt{3-x}}$ , which is defined on the interval $(-1,3)$ . There is one solution to $f'(x)=0$ , namely, $x=2$ . That gives us 3 points to check: the points corresponding to $x=-1,2,3$ . The minimum = 2, which occurs at $x=-1,3$ . The maximum = $2\sqrt{2}$ , which occurs at $x=2$ .

The Minimum value of this function is at x=-1 and x=3 where f(X) becomes 2 and maximum when x=1 where f(x) becomes 2 *2^1/2 therefore a+b+c=6