Question

Find all values of \(c\) such that \(f\) is continuous on \((-\infty, \infty)\). \(f(x)=\left\\{\begin{array}{ll}1-x^{2}, & x \leq c \\ x, & x>c\end{array}\right.\)

Short answer

The values of \(c\) that make \(f(x)\) continuous over all real numbers are \(c = {-1+\sqrt{5}\over 2}\) and \(c = {-1-\sqrt{5}\over 2}\)

Step-by-step solution

Write down the function

The function \(f(x)\) is given as a piecewise function: \(f(x) = 1 - x^2\) for \(x \leq c\) and \(f(x) = x\) for \(x>c\)

Establish the condition for continuity at \(x=c\)

To ensure the function is continuous at \(x=c\), the function values should coincide at \(x=c\) from both sides. In mathematical expression, \(\lim_{x\to c^-} f(x) = \lim_{x\to c^+} f(x)\) which simplifies into \(1 - c^2 = c\)

Solve the equation

Solving the equation \(1 - c^2 = c\) we have to add \(c^2\) to both sides and subtract \(c\) on both sides which gives \(c^2 + c - 1 = 0\)

Solve for \(c\)

To solve for \(c\), use the quadratic formula, \(c={-b\pm\sqrt{b^2-4ac}\over 2a}\), where \(a=1\), \(b=1\), and \(c=-1\). Applying these values, we find \(c = {-1\pm\sqrt{(1)^2-4(1)(-1)}\over 2(1)} = {-1\pm\sqrt{5}\over 2}\)