Answer
C = $\frac{-1 +/- \sqrt 5}{2}$
Work Step by Step
In order to find the values for which a piece wise function is continuous we must check the x value (here it is c) approaching from the right and left. For this function this gets us $\lim\limits_{x \to c^{-}}$f(x) = 1 - $c^{2}$ and f(x) = $\lim\limits_{x \to c^{+}}$f(x) = c. We then set these two equal to one another, 1 - $c^{2}$ = c, and get $c^{2}$ + c - 1 = 0. We can then use the quadratic equation, x = $\frac{-b+/-\sqrt (b^{2} - 4ac)}{2a}$ from the formula a$x^{2}$ + bx + c = 0, plugging in our values from $c^{2}$ + c - 1 = 0 and simplifying, to get c = $\frac{-1 +/- \sqrt 5}{2}$ (note: the c in the formula for the quadratic equation is not the same c as given as an x value in the problem)
