\(R\) is a constant linear quadratic function, which means its natural domain is all real numbers.

\(R(w)\) doesn’t factor easily. factor. We’ll use the quadratic formula.

\(2w^2 - w - 2 = \)

\(a = \answer {2}\)

\(b = \answer {-1}\)

\(c = \answer {-2}\)

\(R\) has two real zeros.

\(R\) is a constant linear quadratic function, which means it is continuous.

The leading coefficient of \(R\) is \(\answer {2}\).

Because \(R\) is a quadratic function with a positive leading coefficient,

\(\lim \limits _{w \to -\infty } R(w) = \answer {\infty }\)

\(\lim \limits _{w \to \infty } R(w) = \answer {\infty }\)

The midpoint between \(\frac {1 + \sqrt {17}}{4} \) and \(\frac {1 - \sqrt {17}}{4}\) is \(\frac {1}{4}\).

Because \(R\) is a quadratic function with a positive leading coefficient, \(R\) decreases on \(\left ( \answer {-\infty }, \answer {\frac {1}{4}} \right ]\) and increases on \(\left [ \answer {\frac {1}{4}}, \answer {\infty } \right )\).

Because \(R\) decreases on \(\left ( -\infty , \frac {1}{4} \right ]\) and increases on \(\left [ \frac {1}{4}, \infty \right )\), \(R\) has a global minimum maximum of \(\answer {-\frac {17}{8}}\) at \(\answer {\frac {1}{4}}\).

Because \(\lim \limits _{w \to \infty } R(w) = \answer {\infty }\), \(R\) has no global minimum maximum.

The only local minimum value of \(R\) is the global minimum of \(\answer {-\frac {17}{8}}\) at \(\answer {\frac {1}{4}}\).

\(R\) is continuous with no maximum value and a minimum value of \(-\frac {17}{8}\). Therefore, the range of \(R\) is