Answer
(a) The function $f(x)$ is continuous for all real numbers. $f(0) \lt 0$ and $f(1) \gt 0$. By the Intermediate Value Theorem, there is a number $c$ in the interval $(0, 1)$ such that $f(c) = 0$
(b) $x = 0.520~~$ is a root of the equation
Work Step by Step
(a) Let $f(x) = arctan~x-1+x$
When $x = 0$:
$f(0) = arctan~0-1+0 = -1$
When $x = 1$:
$f(1) = arctan~1-1+1 = 0.785$
The function $f(x)$ is continuous for all real numbers. $f(0) \lt 0$ and $f(1) \gt 0$. By the Intermediate Value Theorem, there is a number $c$ in the interval $(0, 1)$ such that $f(c) = 0$
(b) Let $f(x) = arctan~x-1+x$
We can use the zoom function on a graphing calculator to see that $f(x) = 0$ when $x \approx 0.520$
