Question 85
Consider the function
$$
f(x) = \frac{x^{2} - 2006x + 1}{x^{2} + 1}
$$
a. Solve the inequality $f'(x) \geq 0$.
b. Prove that $|f(x) - f(y)| \leq 2006$ for all real numbers $x$ and $y$.
$$
f(x) = \frac{x^{2} - 2006x + 1}{x^{2} + 1}
$$
a. Solve the inequality $f'(x) \geq 0$.
b. Prove that $|f(x) - f(y)| \leq 2006$ for all real numbers $x$ and $y$.