Consider the functions $f(x) = \ln\!\left(x\right)$ and $g(x) = x-1$. These are continuous and differentiable for $x > 0$. In this problem we use the Racetrack Principle to show that one of these functions is greater than the other, except at one point where they are equal.

(a) Find a point $c$ such that $f(c) = g(c)$.   $c =$

(b) Find the equation of the tangent line to $f(x) = \ln\!\left(x\right)$ at $x = c$ for the value of $c$ that you found in (a).
$y =$

(c) Based on your work in (a) and (b), what can you say about the derivatives of $f$ and $g$?
$f'(x)$ ? < = > $g'(x)$ for $0 < x < c$, and
$f'(x)$ ? < = > $g'(x)$ for $c < x < \infty$.

(d) Therefore, the Racetrack Principle gives
$f(x)$ ? <= = >= $g(x)$ for $x \le c$, and
$f(x)$ ? <= = >= $g(x)$ for $x \ge c$.